--- title: "Homological Algebra Problem Sets" subtitle: "Problem Set 2" author: - name: D. Zack Garza affiliation: University of Georgia email: dzackgarza@gmail.com date: Spring 2021 order: 2 --- # Tuesday, January 26 :::{.problem title="Weibel 1.3.3"} Prove the 5-lemma. Suppose the following rows are exact: \begin{tikzcd} {A_1} && {A_2} && {A_3} && {A_4} && {A_5} \\ \\ {B_1} && {B_2} && {B_3} && {B_4} && {B_5} \arrow[from=1-1, to=1-3, "{\bd_1^A}"] \arrow[from=1-3, to=1-5, "{\bd_2^A}"] \arrow[from=1-5, to=1-7, "{\bd_3^A}"] \arrow[from=1-7, to=1-9, "{\bd_4^A}"] \arrow[from=3-1, to=3-3, "{\bd_1^B}"] \arrow[from=3-3, to=3-5, "{\bd_2^B}"] \arrow[from=3-5, to=3-7, "{\bd_3^B}"] \arrow[from=3-7, to=3-9, "{\bd_4^B}"] \arrow["{f_1}"{description}, from=1-1, to=3-1] \arrow["{f_2}"{description}, from=1-3, to=3-3] \arrow["{f_3}"{description}, from=1-5, to=3-5] \arrow["{f_4}"{description}, from=1-7, to=3-7] \arrow["{f_5}"{description}, from=1-9, to=3-9] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzAsMCwiQV8xIl0sWzIsMCwiQV8yIl0sWzQsMCwiQV8zIl0sWzYsMCwiQV80Il0sWzgsMCwiQV81Il0sWzAsMiwiQl8xIl0sWzIsMiwiQl8yIl0sWzQsMiwiQl8zIl0sWzYsMiwiQl80Il0sWzgsMiwiQl81Il0sWzAsMV0sWzEsMl0sWzIsM10sWzMsNF0sWzUsNl0sWzYsN10sWzcsOF0sWzgsOV0sWzAsNSwiZl8xIiwxXSxbMSw2LCJmXzIiLDFdLFsyLDcsImZfMyIsMV0sWzMsOCwiZl80IiwxXSxbNCw5LCJmXzUiLDFdXQ==) a. Show that if $f_2, f_4$ are monic and $f_1$ is an epi, then $f_3$ is monic. b. Show that if $f_2, f_4$ are epi and $f_5$ is monic, then $f_3$ is an epi. c. Conclude that if $f_1, f_2, f_4, f_5$ are isomorphisms then $f_4$ is an isomorphism. ::: :::{.solution title="Part (a)"} Appealing to the Freyd-Mitchell Embedding Theorem, we proceed by chasing elements. For this part of the result, only the following portion of the diagram is relevant, where monics have been labeled with "$\injects$" and the epis with "$\surjects$": \begin{tikzcd} {A_1} && {A_2} && {A_3} && {A_4} \\ \\ {B_1} && {B_2} && {B_3} && {B_4} \arrow["{\bd_1^A}", from=1-1, to=1-3] \arrow["{\bd_2^A}", from=1-3, to=1-5] \arrow["{\bd_3^A}", from=1-5, to=1-7] \arrow["{\bd_1^B}", from=3-1, to=3-3] \arrow["{\bd_2^B}", from=3-3, to=3-5] \arrow["{\bd_3^B}", from=3-5, to=3-7] \arrow["{f_1}"{description}, two heads, from=1-1, to=3-1] \arrow["{f_2}"{description}, hook, from=1-3, to=3-3] \arrow["{f_3}"{description}, color={rgb,255:red,92;green,214;blue,92}, from=1-5, to=3-5] \arrow["{f_4}"{description}, hook, from=1-7, to=3-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) It suffices to show that $f_3$ is an injection, and since these can be thought of as \(R\dash\)module morphisms, it further suffices to show that $\ker f_3 = 0$, so $f_3(x) = 0 \implies x = 0$. The following outlines the steps of the diagram chase, with references to specific square and element names in a diagram that follows: - Suppose $x\in A_3$ and $f(x) = 0\in B_3$. - Then under $A_3 \to B_3 \to B_4$, $x$ maps to zero. - Letting $y_1$ be the image of $x$ under $A_3 \to A_4$, commutativity of square 1 and injectivity of $f_4$ forces $y_1 = 0$. - Exactness of the top row allows pulling this back to some $y_2 \in A_2$. - Under $A_2 \to B_2$, $y_2$ maps to some unique $y_3\in B_2$, using injectivity of $f_2$. - Commutativity of square 2 forces $y_3 \to 0$ under $B_2 \to B_3$. - Exactness of the bottom row allows pulling this back to some $y_3\in B_1$ - Surjectivity of $f_1$ allows pulling this back to some $y_5 \in A_1$. - Commutativity of square 3 yields $y_5\mapsto y_2$ under $A_1\to A_2$ and $y_5 \mapsto x$ under $A_1 \to A_2 \to A_3$. - But exactness in the top row forces $y_5 \mapsto 0$ under $A_1 \to A_2 \to A_3$, so $x=0$. \begin{tikzcd} {y_5 \in A_1} && {y_2 \in A_2} && {x\in A_3} && {y_1\in A_4} \\ & 3 && 2 && 1 \\ {y_4 \in B_1} && {y_3 \in B_2} && {0\in B_3} && {0\in B_4} \arrow["{\bd_1^A}", from=1-1, to=1-3] \arrow["{\bd_2^A}", from=1-3, to=1-5] \arrow["{\bd_3^A}", from=1-5, to=1-7] \arrow["{\bd_1^B}", from=3-1, to=3-3] \arrow["{\bd_2^B}", from=3-3, to=3-5] \arrow["{\bd_3^B}", from=3-5, to=3-7] \arrow["{f_1}"{description}, two heads, from=1-1, to=3-1] \arrow["{f_2}"{description}, hook, from=1-3, to=3-3] \arrow["{f_3}"{description}, color={rgb,255:red,92;green,214;blue,92}, from=1-5, to=3-5] \arrow["{f_4}"{description}, hook, from=1-7, to=3-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.solution title="Part (b)"} Similarly, it suffices to consider the following portion of the diagram: \begin{tikzcd} {A_2} && {A_3} && {A_4} && {A_5} \\ \\ {B_2} && {B_3} && {B_4} && {B_5} \arrow["{\bd_2^A}", from=1-1, to=1-3] \arrow["{\bd_3^A}", from=1-3, to=1-5] \arrow["{\bd_4^A}", from=1-5, to=1-7] \arrow["{\bd_2^B}", from=3-1, to=3-3] \arrow["{\bd_3^B}", from=3-3, to=3-5] \arrow["{\bd_4^B}", from=3-5, to=3-7] \arrow["{f_2}"{description}, two heads, from=1-1, to=3-1] \arrow["{f_3}"{description}, color={rgb,255:red,92;green,214;blue,92}, from=1-3, to=3-3] \arrow["{f_4}"{description}, two heads, from=1-5, to=3-5] \arrow["{f_5}"{description}, hook, from=1-7, to=3-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) We'll proceed by starting with an element in $B_3$ and constructing an element in $A_3$ that maps to it. We'll complete this in two steps: first by tracing around the RHS rectangle with corners $B_3, B_5, A_5, A_3$ to produce an "approximation" of a preimage, and second by tracing around the LHS square to produce a "correction term". Various names and relationships between elements are summarized in a diagram following this argument. --- **Step 1** (the right-hand side approximation): - Let ${\color{red}y}\in B_3$ and $y_1$ be its image under $B_3\to B_4$. - By exactness of the bottom row, under $B_4 \to B_5$, $y_1\mapsto 0$. - By surjectivity of $f_4$, pull $y_1$ back to an element $y_2 \in A_4$. - By commutativity of square 1, $y_2\mapsto 0$ under $A_4\to A_5 \to B_5$. - By injectivity of $f_5$, the preimage of zero must be zero and thus $y_2\mapsto 0$ under $A_4\to A_5$. - Using exactness of the top row, pull $y_2$ back to obtain some ${\color{blue}y_3} \in A_3$ **Step 2** (the left-hand correction term): - Let $z$ be the image of $\color{blue}y_3$ under $A_3\to B_3$, noting that $z\neq y$ in general. - By commutativity of square 2, $z\mapsto y_1$ under $B_3\to B_4$ - Thus $z-y\mapsto y_1 - y_1 = 0$ under $B_3\to B_4$, using that $d(z-y) = d(z) - d(y)$ since these are \(R\dash\)module morphisms. - By exactness of the bottom row, pull $z-y$ back to some $\zeta_1\in B_2$. - By surjectivity of $f_2$, pull this back to $\zeta_2 \in A_2$. Note that by construction, $\zeta_2 \mapsto z-y$ under $A_2\to B_2\to B_3$. - Let $\zeta_3$ be the image of $\zeta_2$ under $A_2\to A_3$. - By commutativity of square 3, $\zeta_4 \mapsto z-y$ under $A_3\to B_3$. - But then ${\color{blue}y_3} - \zeta_3 \mapsto z - (z-y) = {\color{red}y}$ under $A_3\to B_3$ as desired. \begin{tikzcd} && \textcolor{rgb,255:red,51;green,112;blue,255}{\zeta_3} \\ \textcolor{rgb,255:red,51;green,112;blue,255}{\zeta_2} && \textcolor{rgb,255:red,153;green,92;blue,214}{y_3} && \textcolor{rgb,255:red,153;green,92;blue,214}{y_2} && \textcolor{rgb,255:red,153;green,92;blue,214}{0} \\ {A_2} && {A_3} && {A_4} && {A_5} \\ & 3 && 2 && 1 \\ {B_2} && {B_3} && {B_4} && {B_5} \\ \textcolor{rgb,255:red,51;green,112;blue,255}{\zeta_1} && \textcolor{rgb,255:red,153;green,92;blue,214}{y} && \textcolor{rgb,255:red,153;green,92;blue,214}{y_1} && \textcolor{rgb,255:red,153;green,92;blue,214}{0} \\ && \textcolor{rgb,255:red,153;green,92;blue,214}{z} \\ && \textcolor{rgb,255:red,51;green,112;blue,255}{z-y} \arrow["{\bd_2^A}", from=3-1, to=3-3] \arrow["{\bd_3^A}", from=3-3, to=3-5] \arrow["{\bd_4^A}", from=3-5, to=3-7] \arrow["{\bd_2^B}", from=5-1, to=5-3] \arrow["{\bd_3^B}", from=5-3, to=5-5] \arrow["{\bd_4^B}", from=5-5, to=5-7] \arrow["{f_2}"{description}, two heads, from=3-1, to=5-1] \arrow["{f_3}"{description}, color={rgb,255:red,92;green,214;blue,92}, from=3-3, to=5-3] \arrow["{f_4}"{description}, two heads, from=3-5, to=5-5] \arrow["{f_5}"{description}, hook, from=3-7, to=5-7] \arrow[color={rgb,255:red,153;green,92;blue,214}, curve={height=-24pt}, dashed, from=2-3, to=7-3] \arrow[color={rgb,255:red,153;green,92;blue,214}, dotted, from=6-3, to=6-5] \arrow[color={rgb,255:red,153;green,92;blue,214}, curve={height=-18pt}, dotted, from=2-5, to=6-5] \arrow[color={rgb,255:red,153;green,92;blue,214}, dotted, from=2-5, to=2-7] \arrow[color={rgb,255:red,153;green,92;blue,214}, dotted, from=2-3, to=2-5] \arrow[color={rgb,255:red,51;green,112;blue,255}, dotted, from=2-1, to=1-3] \arrow[color={rgb,255:red,51;green,112;blue,255}, curve={height=18pt}, dotted, from=2-1, to=6-1] \arrow[color={rgb,255:red,51;green,112;blue,255}, dotted, from=6-1, to=8-3] \arrow[color={rgb,255:red,153;green,92;blue,214}, dotted, from=6-5, to=6-7] \arrow[color={rgb,255:red,153;green,92;blue,214}, curve={height=-12pt}, dotted, from=2-7, to=6-7] \arrow[color={rgb,255:red,51;green,112;blue,255}, curve={height=18pt}, dashed, from=1-3, to=8-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.solution title="Part (c)"} Given the previous two result, if the outer maps are isomorphisms then $f_3$ is both monic and epi. Using a technical fact that monic epis are isomorphisms in a category \( \mathcal{C}\) if and only if \( \mathcal{C} \) is *balanced* and that all abelian categories are balanced, $f_3$ is isomorphism. ::: :::{.problem title="Weibel 1.4.2"} Let $C$ be a chain complex. Show that $C$ is split if and only if there are \(R\dash\)module decompositions \[ C_n & \cong Z_n \oplus B_n' \\ Z_n &= B_n \oplus H_n' .\] Show that $C$ is split exact if and only if $H_n' = 0$. ::: :::{.solution} For this problem, we'll use the fact that if $d$ is an epimorphism, it satisfies the right-cancellation property: if $f\circ d = g\circ d$, then $f=g$. In particular, if $d_n = d_ns_{n-1}d_n$ with $d_n: C_n\to C_{n-1}$ surjective and $s_{n-1}: C_{n-1}\to C_n$, we can conclude $\one_{C_n} = d_ns_{n-1}$. We'll also use the fact that if we have a SES in any abelian category \( \mathcal{A} \), then the following are equivalent: - The sequence is split on the left. - The sequence is split on the right. - The middle term is isomorphic to the direct sum of the outer terms. Fixing notation, we'll write $C \da (\cdots \to C_{n+1} \mapsvia{\bd_{n+1}} C_n \mapsvia{\bd_n} \cdots)$, and we'll use concatenation $fg$ to denote function composition $f \circ g$. $\implies$: Suppose $C$ is split, so we have maps $\ts{s_n}$ such that $\bd_n = \bd_n s_{n-1} \bd_n$. :::{.claim} The short exact sequence \[ 0 \to Z_n \mapsvia{\iota} C_n \mapsvia{\bd_n} B_{n-1} \to 0 \] admits a right-splitting $f: B_{n-1}\to C_n$, and thus there is an isomorphism \[ C_n \cong Z_n \oplus B_n' \da Z_n \oplus B_{n-1} .\] ::: :::{.proof title="?"} That this sequence is exact follows from the fact that it can be written as \[ 0 \to \ker \bd_n \injects C_n \overset{\bd_n}\surjects \im \bd _n \to 0 .\] We proceed by constructing the splitting $f$. Noting that $s_{n-1}: C_{n-1} \to C_n$ and $B_{n-1} \leq C_{n-1}$, the claim is that its restriction $f\da \ro{s_{n-1}}{B_{n-1}}$ works. It suffices to show that $(C_n \mapsvia{\bd_n} B_{n-1} \mapsvia{f} C_n)$ composes to the identity map $\one_{C_n}$. This follows from the splitting assumption, along with right-cancellability since $\bd_n$ is surjective onto its image: \[ \bd_n = \bd_n s_{n-1} \bd _n \overset{\text{right-cancel $\bd_n$}}{\implies} \one_{C_n} = \bd_n s_{n-1} \da \bd_n f .\] ::: :::{.claim} The SES \[ 0 \to B_n \overset{\iota_{BZ}}\injects Z_n \overset{\pi}{\surjects} \frac{Z_n}{B_n} \to 0 \] admits a left-splitting $f: Z_n\to B_n$, and thus there is an isomorphism \[ Z_n \cong B_n \oplus H_n' \da B_n \oplus H_n(C) \da B_n \oplus {Z_n \over B_n} .\] ::: :::{.proof title="?"} We proceed by again constructing the splitting $f:Z_n\to B_n$. The situation is summarized in the following diagram: \begin{tikzcd} \cdots && {C_{n+1}} & {} & {C_n} && \cdots \\ \\ &&&& {Z_n} \\ \\ &&&& {B_n} \arrow[from=1-1, to=1-3] \arrow["{\bd_{n+1}}", from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow["{\iota_{BZ}}"', hook', from=5-5, to=3-5] \arrow["{\iota_{Z}}"', hook', from=3-5, to=1-5] \arrow["{s_{n}}"', curve={height=18pt}, dashed, from=1-5, to=1-3] \arrow["{\bd_{n+1}}"', two heads, from=1-3, to=5-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) So a natural candidate for the map $f$ is the composition \[ Z_n \mapsvia{\iota_Z} C_n \mapsvia{s_{n}} C_{n+1} \mapsvia{\bd_{n+1}} B_n ,\] so $f \da \bd_{n+1} s_{n} \iota_Z$. We can simplify this slightly by regarding $Z_n \leq C_n$ as a submodule to suppress $\iota_Z$, and identify $s_{n}$ with its restriction to $Z_n$ to write $f \da \bd_{n+1} s_{n}$. The claim is then that $f\iota_{BZ} = \one_{B_n}$. Anticipating using the fact that $\bd_{n+1} = \bd_{n+1} s_n \bd_{n+1}$, we post-compose with $\bd_{n+1}$ and compute: \[ f \iota_{BZ} \bd_{n+1} &= (C_{n+1} \mapsvia{\bd_{n+1}} B_n \mapsvia{\iota_{BZ}} Z_n \mapsvia{\iota_Z} C_n \mapsvia{s_n} C_{n+1} \mapsvia{\bd_{n+1}} C_n ) \\ &= (C_{n+1} \mapsvia{\bd_{n+1}} B_n \mapsvia{\ro{s_n}{B_n} } C_{n+1} \mapsvia{\bd_{n+1}} ) \\ &= \bd_{n+1} s_n \bd_{n+1} \\ &= \bd_{n+1} ,\] where in the last step we've used the splitting hypothesis and the fact that it remains true when everything is restricted to the submodule $B_n \leq C_n$. Using surjectivity of $\bd_{n+1}$ onto $B_n$, we can now conclude as before: \[ f \iota_{BZ} \bd_{n+1} = \bd_{n+1} \overset{\text{right-cancel $\bd_n$}}{\implies} f\iota_{BZ} = \one_{B_n} .\] ::: ::: :::{.problem title="Weibel 1.4.3"} Show that $C$ is a split exact chain complex if and only if $\one_C$ is nullhomotopic. ::: :::{.solution} $\impliedby$: **$C$ is split**: Suppose $\one_{C}$ is nullhomotopic, so that there exist maps \[ s_n: C_n \to C_{n+1} && \one_{C_n} = \bd_{n+1} s_n + s_{n-1} \bd_n .\] We then have the following situation: \begin{tikzcd} \cdots && {C_{n+1}} && {C_{n}} && {C_{n-1}} && \cdots \\ \\ \cdots && {C_{n+1}} && {C_{n}} && {C_{n-1}} && \cdots \arrow["{\bd_{n+1}}", color={rgb,255:red,92;green,214;blue,92}, from=1-3, to=1-5] \arrow["{\bd_{n+1}}", color={rgb,255:red,214;green,92;blue,92}, from=3-3, to=3-5] \arrow[from=1-1, to=1-3] \arrow[from=3-1, to=3-3] \arrow["{\bd_{n}}", color={rgb,255:red,214;green,92;blue,92}, from=1-5, to=1-7] \arrow["{\bd_{n}}", from=3-5, to=3-7] \arrow["{\one_{C_n}}", color={rgb,255:red,92;green,214;blue,92}, from=1-5, to=3-5] \arrow["{\one_{C_{n+1}}}"', from=1-3, to=3-3] \arrow["{s_n}", color={rgb,255:red,214;green,92;blue,92}, from=1-5, to=3-3] \arrow["{s_n}"{description}, curve={height=24pt}, dashed, from=1-5, to=1-3] \arrow[from=1-7, to=1-9] \arrow[from=3-7, to=3-9] \arrow["{\one_{C_{n-1}}}", from=1-7, to=3-7] \arrow["{s_{n-1}}", color={rgb,255:red,214;green,92;blue,92}, from=1-7, to=3-5] \arrow["{s_{n-1}}"{description}, curve={height=24pt}, dashed, from=1-7, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzIsMCwiQ197bisxfSJdLFs0LDAsIkNfe259Il0sWzIsMiwiQ197bisxfSJdLFs0LDIsIkNfe259Il0sWzAsMCwiXFxjZG90cyJdLFswLDIsIlxcY2RvdHMiXSxbNiwwLCJDX3tuLTF9Il0sWzYsMiwiQ197bi0xfSJdLFs4LDAsIlxcY2RvdHMiXSxbOCwyLCJcXGNkb3RzIl0sWzAsMSwiXFxiZF97bisxfSIsMCx7ImNvbG91ciI6WzEyMCw2MCw2MF19LFsxMjAsNjAsNjAsMV1dLFsyLDMsIlxcYmRfe24rMX0iLDAseyJjb2xvdXIiOlswLDYwLDYwXX0sWzAsNjAsNjAsMV1dLFs0LDBdLFs1LDJdLFsxLDYsIlxcYmRfe259IiwwLHsiY29sb3VyIjpbMCw2MCw2MF19LFswLDYwLDYwLDFdXSxbMyw3LCJcXGJkX3tufSJdLFsxLDMsIlxcb25lX3tDX259IiwwLHsiY29sb3VyIjpbMTIwLDYwLDYwXX0sWzEyMCw2MCw2MCwxXV0sWzAsMiwiXFxvbmVfe0Nfe24rMX19IiwyXSxbMSwyLCJzX24iLDAseyJjb2xvdXIiOlswLDYwLDYwXX0sWzAsNjAsNjAsMV1dLFsxLDAsInNfbiIsMSx7ImN1cnZlIjo0LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNiw4XSxbNyw5XSxbNiw3LCJcXG9uZV97Q197bi0xfX0iXSxbNiwzLCJzX3tuLTF9IiwwLHsiY29sb3VyIjpbMCw2MCw2MF19LFswLDYwLDYwLDFdXSxbNiwxLCJzX3tuLTF9IiwxLHsiY3VydmUiOjQsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) Here the nullhomotopy is outlined in red, and the map relevant to the splitting in green. Note that $s_n: C_n \to C_{n+1}$ is a candidate for a splitting, we just need to show that $\bd_{n+1} = \bd_{n+1} s_n \bd_{n+1}$. We can proceed by post-composing the LHS with the identity $\one_C$, which allows us to substitute in the nullhomotopy: \[ \bd_{n+1} &= \one_{C_n} \bd_{n+1} \\ &= \qty{ \bd_{n+1} s_n + s_{n-1} \bd_n } \bd_{n+1} \\ &= \bd_{n+1} s_n \bd_{n+1} + s_{n-1} \bd_n \bd_{n+1} \\ &= \bd_{n+1} s_n \bd_{n+1} + s_{n-1} {\color{red} 0} && \text{since } \bd^2 = 0\\ &= \bd_{n+1} s_n \bd_{n+1} .\] **$C$ is exact**: This follows from the fact that since $\one_C = \bd s + s\bd$ are equal as maps of chain complexes, the images $D_1 \da \one_C(C)$ and $D_2 \da (\bd s + s \bd)(C)$ are equal as chain complexes and have equal homology. Evidently $D_1 = C$, and on the other hand, each graded piece $(D_2)_n$ *only* consists of boundaries coming from various pieces of $C$, since $\bd s + s\bd$ necessarily lands in the images of the maps $\bd_n$. Thus $C_n(D_2) \subseteq B_n(D_2) = \emptyset$, i.e. there are no chains (or cycles) in $D_2$ which are *not* boundaries, and thus $H_n(D_2) \da Z_n(D_2)/ B_n(D_2) = 0$ for all $n$. We can thus conclude that $C = D_2 \implies H(C) = H(D_2) = 0$, so $C$ must be exact. --- $\implies$: Suppose $C$ is split. Then by exercise 1.4.2, we have \(R\dash\)module decompositions \[ C_n &\cong Z_n \oplus B_{n-1} \\ Z_n &\cong B_n \oplus H_n \\ \\ \implies C_n &\cong B_n \oplus B_{n-1} \oplus H_n .\] Supposing further that $C$ is exact, we have $H_n = 0$, and thus $C_n \cong B_n \oplus B_{n-1}$. We first note that in this case, we can explicitly write the differential $\bd_n$. Letting $\one_n$ denote the identity on $C_n$, where by abuse of notation we also write this for its restriction to any submodules, we have: \begin{tikzcd} {C_n} && {C_{n-1}} \\ \\ {B_n} && {B_{n-1}} \\ \oplus && \oplus \\ {B_{n-1}} && {B_{n-2}} \arrow["{\bd_n}", from=1-1, to=1-3] \arrow["{=}"{marking}, draw=none, from=1-1, to=3-1] \arrow["{=}"{marking}, draw=none, from=1-3, to=3-3] \arrow["{\one_{n}}"{description, pos=0.3}, from=5-1, to=3-3] \arrow["0"{description, pos=0.2}, dashed, from=3-1, to=3-3] \arrow["0"{description, pos=0.2}, dashed, from=3-1, to=5-3] \arrow["0"{description, pos=0.2}, dashed, from=5-1, to=5-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) We can thus write $\bd_n$ as the matrix \[ \bd_n = \begin{bmatrix} 0 & \one_n \\ 0 & 0 \end{bmatrix} .\] Similarly using this decomposition, we can construct a map $s_n: C_{n} \to C_{n+1}$: \begin{tikzcd} {C_n} && {C_{n+1}} \\ \\ {B_n} && {B_{n+1}} \\ \oplus && \oplus \\ {B_{n-1}} && {B_{n}} \arrow["{s_n}", from=1-1, to=1-3] \arrow["{=}"{marking}, draw=none, from=1-1, to=3-1] \arrow["{=}"{marking}, draw=none, from=1-3, to=3-3] \arrow["{\one_n}"{description, pos=0.3}, from=3-1, to=5-3] \arrow["0"{description, pos=0.2}, dashed, from=3-1, to=3-3] \arrow["0"{description, pos=0.2}, dashed, from=5-1, to=5-3] \arrow["0"{description, pos=0.2}, dashed, from=5-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) We can write this as the following matrix: \[ s_n = \begin{bmatrix} 0 & 0 \\ \one_n & 0 \end{bmatrix} .\] We can now verify that $s_n$ is a nullhomotopy from a direct computation: \[ \bd_{n+1} s_n + s_{n-1} \bd_n &= \begin{bmatrix} 0 & \one_{n+1} \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ \one_n & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ \one_{n-1} & 0 \end{bmatrix} \begin{bmatrix} 0 & \one_{n} \\ 0 & 0 \end{bmatrix} \\ &= \begin{bmatrix} \one_n & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & \one_{n-1} \end{bmatrix} \\ &= \begin{bmatrix} \one_{B_n} & 0 \\ 0 & \one_{B_{n-1}} \end{bmatrix} \\ &= \one_{C_n} ,\] expressed as a map $B_n \oplus B_{n-1} \to B_n \oplus B_{n-1}$. ::: :::{.problem title="Weibel 1.4.5"} Show that chain homotopy classes of maps form a quotient category $K$ of $\Ch(\rmod)$ and that the functors $H_n$ factor through the quotient functor $\Ch(\rmod) \to K$ using the following steps: 1. Show that chain homotopy equivalence is an equivalence relation on $\ts{f:C\to D \st f\text{ is a chain map}}$. Define $\Hom_K(C, D)$ to be the equivalence classes of such maps and show that it is an abelian group. 2. Let $f \homotopic g: C\to D$ be two chain homotopic maps. If $u: B\to C, v:D\to E$ are chain maps, show that $vfu, vgu$ are chain homotopic. Deduce that $K$ is a category when the objects are defined as chain complexes and the morphisms are defined as in (1). 3. Let $f_0, f_1, g_0, g_1: C\to D$ all be chain maps such that each pair $f_i \homotopic g_i$ are chain homotopic. Show that $f_0 + f_1$ is chain homotopic to $g_0 + g_1$. Deduce that $K$ is an additive category and $\Ch(\rmod)\to K$ is an additive functor. 4. Is $K$ an abelian category? Explain. > Try at least two parts. ::: :::{.solution title="Part 1"} \envlist :::{.claim title="1"} Chain homotopy equivalence defines an equivalence relation on $\Hom_{\Ch(\mathcal{A})}(A, B)$. ::: :::{.proof title="of claim 1"} We recall that for morphisms $f,g \in \Hom_{\Ch(\mathcal{A} )}(A, B)$, we have $f \homotopic g \iff f-g \homotopic 0 \iff \exists s: A \to B[1]$ such that $\bd^B s + s\bd^A = f-g$. - **Reflexive:** We want to show that $f \homotopic f$, i.e. $f-f = 0 \homotopic 0$. Producing the map $s = 0$ works, since $\bd s + s\bd = \bd 0 + 0\bd = 0$. - **Symmetric**: Suppose $f\homotopic g$, so there exists an $s$ such that $\bd s + s \bd = f-g$. Then taking $s' \da -s$ produces a chain homotopy $g-f\homotopic 0$, since we can write \[ \bd s' + s' \bd &= \bd(-s) + (-s) \bd \\ &= -\bd s - s\bd \\ &= -( \bd s + s \bd) \\ &= -(f-g) \\ &= g-f .\] - **Transitive**: Suppose $f\homotopic g$ and $g\homotopic h$, we want to show $f\homotopic h$. By assumption we have maps $s, s'$ such that $\bd s + s\bd = f-g$ and $\bd s' + s'\bd = g-h$, so set $s'' \da s + s'$. We can then check that this is a chain homotopy from $f$ to $h$: \[ \bd s'' + s'' \bd &= \bd(s+ s') + (s + s')\bd \\ &= (\bd s + s\bd) + (\bd s' + s' \bd) \\ &= (f-g) + (g-h) \\ &= f-h ,\] where we've used that abelian categories are enriched over abelian groups, so we have a commutative and associative addition on homs. ::: :::{.claim title="2"} $( \Hom_K(A, B), \oplus) \in \Ab$, where we define an addition on equivalence classes by \[ [f] \oplus [g] \da [f + g] .\] ::: :::{.proof title="of claim 2"} We check the group axioms directly: - **Closure under operation**: We can check that $[f] \oplus [g] \da [f+g] \da [g']$ makes sense, since $g' \da f+g \in \Hom_{\Ch(\mathcal{A})}(A, B)$, making $[g']$ a well-defined equivalence class of maps in $\Hom_K(A, B)$. - **Two-sided Identities**: The equivalence class $[0] \da \ts{f\in \Hom_{\Ch(\mathcal{A} )}(A, B) \st f\homotopic 0}$ serves as an identity, where we take the zero map as a representative. This follows from the fact that \[ [0] \oplus [f] \da [0 + f] = [f] = [f] \oplus [0] .\] - **Associativity**: This again follows from the abelian group structure on the original hom: \[ [f] \oplus \qty{ [g] \oplus [h] } = [f + (g+h)] = [(f+g) + h] = \qty{ [f] \oplus [g] } \oplus [h] .\] - **Two-sided inverses**: Let $\ominus[f] \da [-f]$ be a candidate for the inverse with respect to $\oplus$. To see that this works, we have \[ [f] \oplus ( \ominus [f] ) \da [f] \oplus [-f] = [f-f] = [0] ,\] and a similar calculation goes through to show it's also a left-sided inverse. - **Well-definedness**: We need to show that if $[f] = [f']$ and $[g] = [g']$, the sums agree, so $[f] \oplus [g] = [f'] \oplus [g']$. We'll use the fact that $[f] = [f'] \iff f-f' \homotopic 0$ (and similarly for $g$), so we can compute: \[ \qty{ [f] \oplus [g] } \ominus \qty{ [f'] \oplus [g'] } &\da [f + g] \ominus [f' - g'] \\ &\da [(f+g) - (f'-g') ] \\ &= [(f-f') + (g-g')] \\ &= [f-f'] \oplus [g-g'] \\ &= [0] \oplus [0] && \text{using } f\homotopic f',~ g\homotopic g'\\ &\da [0] .\] ::: ::: :::{.solution title="Part 2"} We have the following situation, where the double-arrows denote chain homotopies: \begin{tikzcd} C &&& D \\ \\ B &&& E \arrow[""{name=0, anchor=center, inner sep=0}, "f"{description}, curve={height=-18pt}, from=1-1, to=1-4] \arrow[""{name=1, anchor=center, inner sep=0}, "g"{description}, curve={height=12pt}, from=1-1, to=1-4] \arrow["u", from=3-1, to=1-1] \arrow["v", from=1-4, to=3-4] \arrow[""{name=2, anchor=center, inner sep=0}, "vfu"{description}, curve={height=-18pt}, from=3-1, to=3-4] \arrow[""{name=3, anchor=center, inner sep=0}, "vgu"{description}, curve={height=18pt}, from=3-1, to=3-4] \arrow[shorten <=4pt, shorten >=4pt, Rightarrow, from=0, to=1] \arrow["\exists"', shorten <=5pt, shorten >=5pt, Rightarrow, dashed, from=2, to=3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJCIl0sWzAsMCwiQyJdLFszLDAsIkQiXSxbMywyLCJFIl0sWzEsMiwiZiIsMSx7ImN1cnZlIjotM31dLFsxLDIsImciLDEseyJjdXJ2ZSI6Mn1dLFswLDEsInUiXSxbMiwzLCJ2Il0sWzAsMywidmZ1IiwxLHsiY3VydmUiOi0zfV0sWzAsMywidmd1IiwxLHsiY3VydmUiOjN9XSxbNCw1LCIiLDAseyJzaG9ydGVuIjp7InNvdXJjZSI6MjAsInRhcmdldCI6MjB9fV0sWzgsOSwiXFxleGlzdHMiLDIseyJzaG9ydGVuIjp7InNvdXJjZSI6MjAsInRhcmdldCI6MjB9LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) We want to show that $vgu \homotopic vfu$, or equivalently that $vgu - vfu \homotopic 0$. This is immediate: \[ vgu - vfu = v(gu - fu) = v(f - g)u \homotopic v0u = 0 .\] Alternatively, as an explicit computation, if we assume $f\homotopic g$ then there is a nullhomotopy $s$ for $f-g$, in which case the map $s' \da vsu$ works as a nullhomotopy for $vfu - vgu$: \[ vfu - vgu &= v(f-g)u = v(\bd s + s \bd) u \\ &= {\color{red} v\bd} s u + vs {\color{red} \bd u } \\ &= {\color{blue} \bd v} su + v s {\color{blue} u \bd } && \text{since $u, v$ are chain maps} \\ &\da \bd s' + s'\bd .\] We can now define a composition map on $\cat{K}$: \[ \circ: \Hom_K(A_1, A_2) \cross \Hom_K(A_2, A_3) &\to \Hom_K(A_1, A_3) \\ ([f], [g]) &\mapsto [f\circ g] .\] The previous argument then precisely says: 1. If $u:B\to C$ and $f,g C\to D$ with $[f] = [g]$, the composition $[u] \circ [f] = [u] \circ [g]: B\to D$ is well-defined, and 2. If $v:D\to E$ and $f,g: C\to D$ with $[f] = [g]$, the composition $[f] \circ [v] = [g] \circ [v]$ is well-defined. So this composition on $\cat{K}$ is well-defined. Moreover, - There is an identity morphism $[\one_A] \in \Hom_K(A, A)$ coming from the class of $\one_C \in \Hom_{\Ch( \mathcal{A} ) }(A, A)$, - It is associative, since $[h] \circ [gf] \da [hgf] \da [hg] \circ [f]$, and - It is unital, in the sense that $[\one_B] \circ [f] \da [\one_A \circ f] = [f] = [f \circ \one_A] \da [f] \circ [\one_B]$ for any $f:A\to B$. So this data satisfies all of the axioms of a category (Weibel A.1.1). ::: :::{.solution title="Part 3"} To see that $f_0 + f_1 \homotopic g_0 + g_1$, we have \[ (f_0 + f_1) - (g_0 + g_1) &= (f_0 - g_0) + ( f_1 - g_1) \\ &\homotopic 0 + 0 = 0 .\] To be explicit, if $s_i$ are chain homotopies for $f_i, g_i$, we can take $s\da s_0 + s_1$: \[ (f_0 + f_1) - (g_0 + g_1) &= (f_0 - g_0) + ( f_1 - g_1) \\ &= (\bd s_0 + s_0 \bd) + (\bd s_1 + s_1 \bd) \\ &= (\bd s_0 + \bd s_1) + (s_0 \bd + s_1 \bd) \\ &= \bd(s_0 + s_1) + (s_0 + s_1)\bd \\ &\da \bd s + s \bd .\] That $K$ forms an additive category is a consequence of the following facts: - $K$ has products, since \( \mathcal{A} \) had products and $\Ob(K) \da \Ob( \mathcal{A})$. - $K$ has a zero object, for the same reason. - Composition distributes over addition, i.e. \[ [f] \qty{ [g] \oplus [h] } \\ &\da [f] [g +h] \\ &\da [f(g + h) ] \\ &\da [fg + fh] \\ &= [fg] \oplus [fh] .\] Moreover, the quotient functor $\Ch(\rmod) \to K$ is an additive functor, since the maps $Q: \Hom_{\Ch( \mathcal{A} )}(A, B) \to \Hom_{\Ch( \mathcal{A} )}(A, B)/ \sim$ are morphisms of abelian groups, using the fact that $Q$ commutes with both additions: \[ Q(f + g) = [f + g] \da [f] \oplus [g] = Q(f) + Q(g) .\] Alternatively, we can note that the set of all $f\in \Hom_{ \Ch( \mathcal{A} }(A, B)$ which are nullhomotopic form a subgroup $H$, and since everything is abelian we can form the quotient $\Hom_{ \Ch(\mathcal{A}) } (A, B) / H$ observe that this is isomorphic as a group to $\Hom_K(A, B)$. ::: :::{.solution title="Part 4"} We can first note that $K$ is an additive category, since we still have a zero object and products inherited from $\Ch( \mathcal{A})$. > Note: I don't see a great way to prove that *any* particular category is abelian or not! > Checking the axioms listed in Appendix A.4 seems quite difficult. ::: :::{.problem title="Weibel 1.5.1"} Let $\cone(C) \da \cone(\one_C)$, so \[ \cone(C)_n = C_{n-1} \oplus C_n .\] Show that $\cone(C)$ is split exact, with splitting map given by $s(b, c) \da (-c, 0)$. ::: :::{.solution} Fixing notation, let - $\bd_n$ be the $n$th differential on $C$, - $\one_C$ be the identity chain map on $C$, - $\one_n: C_n\to C_n$ be the $n$th graded component of $\one_C$, - $\hat{C} \da \cone(C) \da \cone(\one_C)$, - $\hat{\bd}_n$ be the $n$th differential on $\hat{C}$, and - $\hat{\one}$ be the identity on $\hat{C}$, - $\hat{\one}_n: \hat{C}_n \to \hat{C}_n$ be the $n$th component of $\hat{\one}$. From exercise 1.4.2, it suffices to show that $\hat{\one}$ is nullhomotopic. Since we have a direct sum decomposition $\cone(C)_n \da C_{n-1} \oplus C_n$, we can write $\hat{\one}$ as a block matrix \[ \begin{bmatrix} \one_{n-1} & 0 \\ 0 & \one_n \end{bmatrix} .\] We can similarly write down the differential on $\cone(C)$ in block form: \begin{tikzcd} {\cone(C)_{n+1}} &&& {\cone(C)_n} \\ \\ {C_n} &&& {C_{n-1}} \\ \oplus &&& \oplus \\ {C_{n+1}} &&& {C_n} \arrow["{\hat{\bd}_{n+1}}", from=1-1, to=1-4] \arrow["{-\bd_n}"{pos=0.2}, from=3-1, to=3-4] \arrow["{-\one_n}"'{pos=0.2}, from=3-1, to=5-4] \arrow["0"'{pos=0.2}, dashed, from=5-1, to=3-4] \arrow["{\bd_{n+1}}"'{pos=0.3}, from=5-1, to=5-4] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwyLCJDX24iXSxbMCwwLCJcXGNvbmUoQylfe24rMX0iXSxbMywwLCJcXGNvbmUoQylfbiJdLFswLDQsIkNfe24rMX0iXSxbMywyLCJDX3tuLTF9Il0sWzMsNCwiQ19uIl0sWzAsMywiXFxvcGx1cyJdLFszLDMsIlxcb3BsdXMiXSxbMSwyLCJcXGhhdHtcXGJkfV97bisxfSJdLFswLDQsIi1cXGJkX24iLDAseyJsYWJlbF9wb3NpdGlvbiI6MjB9XSxbMCw1LCItXFxvbmVfbiIsMix7ImxhYmVsX3Bvc2l0aW9uIjoyMH1dLFszLDQsIjAiLDIseyJsYWJlbF9wb3NpdGlvbiI6MjAsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDUsIlxcYmRfe24rMX0iLDIseyJsYWJlbF9wb3NpdGlvbiI6MzB9XV0=) This yields \[ \hat{\bd}_n \da \begin{bmatrix} -\bd_{n-1} & 0 \\ -\one_{n-1} & \bd_n \end{bmatrix} .\] Similarly, the map $s_n(b, c) = (-c, 0)$ can be written as \[ s_n = \begin{bmatrix} 0 & -\one_n \\ 0 & 0 \end{bmatrix} .\] We can thus proceed by a direct computation: \[ \bd_{n+1} s_n + s_{n-1} \bd_n &= \begin{bmatrix} -\bd_n & 0 \\ -\one_n & \bd_{n+1} \end{bmatrix} \begin{bmatrix} 0 & -\one_n \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & -\one_{n-1} \\ 0 & 0 \end{bmatrix} \begin{bmatrix} -\bd_{n-1} & 0 \\ -\one_{n-1} & \bd_n \end{bmatrix} \\ \\ &= \begin{bmatrix} 0 & \bd_n \one_n \\ 0 & \one_n \one_n \end{bmatrix} + \begin{bmatrix} \one_{n-1}\one_{n-1} & -\one_{n-1}\bd_n \\ 0 & 0 \end{bmatrix} \\ \\ &= \begin{bmatrix} \one_{n-1} \one_{n-1} & \bd_n \one_n - \one_{n-1} \bd_n \\ 0 & \one_n \one_n \end{bmatrix} \\ \\ &= \begin{bmatrix} \one_{n-1} \one_{n-1} & 0 \\ 0 & \one_n \one_n \end{bmatrix} \\ &= \hat{\one}_n .\] ::: :::{.problem title="Weibel 1.5.2"} Let $f:C\to D \in \Mor(\Ch(\mathcal{A}))$ and show that $f$ is nullhomotopic if and only if $f$ lifts to a map \[ (-s, f): \cone(C) \to D .\] ::: :::{.solution} \envlist :::{.remark} As a notational convention for this problem, I'll take vectors $v$ to be column vectors, and $v^t$ will denote a row vector. I've also written $f \homotopic 0$ to denote that $f$ is nullhomotopic. ::: $\implies$: Suppose that $f \homotopic 0$, so there are maps $s_n$ such that the following diagrams commute for every $n$: \begin{tikzcd} && \cdots && {C_n} && {C_{n-1}} && \cdots \\ \\ \cdots && {D_{n+1}} && {D_n} && \cdots \arrow["{\bd_n^C}", from=1-5, to=1-7] \arrow["{f_n}"', from=1-5, to=3-5] \arrow["{s_n}"{description}, color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-5, to=3-3] \arrow["{\bd^D_{n+1}}"{description}, from=3-3, to=3-5] \arrow["{s_{n-1}}"{description}, color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-7, to=3-5] \arrow[from=1-3, to=1-5] \arrow[from=1-7, to=1-9] \arrow[from=3-5, to=3-7] \arrow[from=3-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbNCwwLCJDX24iXSxbNiwwLCJDX3tuLTF9Il0sWzIsMiwiRF97bisxfSJdLFs0LDIsIkRfbiJdLFsyLDAsIlxcY2RvdHMiXSxbOCwwLCJcXGNkb3RzIl0sWzYsMiwiXFxjZG90cyJdLFswLDIsIlxcY2RvdHMiXSxbMCwxLCJcXGJkX25eQyJdLFswLDMsImZfbiIsMl0sWzAsMiwic19uIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbMjQwLDYwLDYwLDFdXSxbMiwzLCJcXGJkXkRfe24rMX0iLDFdLFsxLDMsInNfe24tMX0iLDEseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19LFsyNDAsNjAsNjAsMV1dLFs0LDBdLFsxLDVdLFszLDZdLFs3LDJdXQ==) Write $\hat{C} \da \cone(C) \da \cone(\one_{C}) \da C[1] \oplus C$, we then want to construct a lift $\hat{f}$ of $f$ such that the following diagram commutes: \begin{tikzcd} {\hat{C}} & {} \\ \\ C && D \arrow["f", from=3-1, to=3-3] \arrow["{{\iota}_{\hat C}}", from=3-1, to=1-1] \arrow["{\hat f}", dashed, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXGhhdHtDfSJdLFswLDIsIkMiXSxbMiwyLCJEIl0sWzEsMF0sWzEsMiwiZiJdLFsxLDAsIntcXGlvdGF9X3tcXGhhdCBDfSJdLFswLDIsIlxcaGF0IGYiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) where $\iota_{\hat C}$ is the following inclusion of $C$ into its cone: \begin{tikzcd} {C_n} &&& {\hat{C}_n} \\ \\ {C_n} &&& {C_{n-1}} \\ &&& \oplus \\ &&& {C_n} \arrow["{\one_{C_n}}", from=3-1, to=5-4] \arrow["0", dashed, from=3-1, to=3-4] \arrow["{\iota_{\hat{C}}}", from=1-1, to=1-4] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJDX24iXSxbMywwLCJcXGhhdHtDfV9uIl0sWzMsMiwiQ197bi0xfSJdLFszLDQsIkNfbiJdLFszLDMsIlxcb3BsdXMiXSxbMCwyLCJDX24iXSxbNSwzLCJcXG9uZV97Q19ufSJdLFs1LDIsIjAiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMCwxLCJcXGlvdGFfe1xcaGF0e0N9fSJdXQ==) We define the map $\hat{f}$ in the following way: \begin{tikzcd} {C_n} &&& {\hat{C}_n} && {D_n} \\ \\ {C_n} &&& {C_{n-1}} \\ &&& \oplus \\ &&& {C_n} && {D_n} \arrow["{\one_{C_n}}", from=3-1, to=5-4] \arrow["0", dashed, from=3-1, to=3-4] \arrow["{\iota_{\hat{C}}}", from=1-1, to=1-4] \arrow["{\hat{f}_n}", color={rgb,255:red,153;green,92;blue,214}, from=1-4, to=1-6] \arrow["{f_n}", color={rgb,255:red,153;green,92;blue,214}, from=5-4, to=5-6] \arrow["{-s_{n-1}}"{description}, color={rgb,255:red,153;green,92;blue,214}, from=3-4, to=5-6] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwwLCJDX24iXSxbMywwLCJcXGhhdHtDfV9uIl0sWzMsMiwiQ197bi0xfSJdLFszLDQsIkNfbiJdLFszLDMsIlxcb3BsdXMiXSxbMCwyLCJDX24iXSxbNSwwLCJEX24iXSxbNSw0LCJEX24iXSxbNSwzLCJcXG9uZV97Q19ufSJdLFs1LDIsIjAiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMCwxLCJcXGlvdGFfe1xcaGF0e0N9fSJdLFsxLDYsIlxcaGF0e2Z9X24iLDAseyJjb2xvdXIiOlsyNzAsNjAsNjBdfSxbMjcwLDYwLDYwLDFdXSxbMyw3LCJmX24iLDAseyJjb2xvdXIiOlsyNzAsNjAsNjBdfSxbMjcwLDYwLDYwLDFdXSxbMiw3LCItc197bi0xfSIsMSx7ImNvbG91ciI6WzI3MCw2MCw2MF19LFsyNzAsNjAsNjAsMV1dXQ==) That this is a lift follows from computing the composition, which can be done in block matrices: \[ \hat{f}_n \circ \iota_{\hat C, n} &= \begin{bmatrix} -s_{n-1} \\ f_n \end{bmatrix} ^t \begin{bmatrix} 0 \\ \one_{C_n} \end{bmatrix} = [ f_n \one_{C_n} ] = f_n ,\] where the first matrix acts as a row vector. It only remains to check that $\hat{f}$ defines a chain map, which follows from the following computation: \[ \bd_n^D \hat{f}_n - \hat{f}_{n-1} \hat{\bd}_n &= [\bd_n^D] \begin{bmatrix} -s_{n-1} \\ f_n \end{bmatrix} ^t - \begin{bmatrix} -s_{n-2} \\ f_{n-1} \end{bmatrix} \begin{bmatrix} -\bd_n^C & 0 \\ -\one_{C_n} & \bd_{n+1}^C \end{bmatrix} \\ \\ &= \begin{bmatrix} \bd_n^D (-s_{n-1}) - s_{n-2}\bd_n^C + f_{n-1} \one_{C_n} \\ \bd_n^D f_n - f_{n-1} \bd_n^C \end{bmatrix} ^t \\ \\ &= \begin{bmatrix} f_{n-1} -( \bd_n^D s_{n-1} + s_{n-2}\bd_n^C) \\ 0 \end{bmatrix} ^t && \text{since $f$ a chain map $\implies \bd f=f \bd$} \\ \\ &= \begin{bmatrix} 0 \\ 0 \end{bmatrix} && \text{since $f \homotopic 0 \implies \bd s + s\bd = f$} \\ &=0 .\] --- $\impliedby$: Suppose we have a lift $\hat f: \hat{C} \to D$; then define the following maps as the proposed splittings: \[ s_n: C_{n-1} &\to D_n \\ c &\mapsto \hat{f}_n(-c, 0) .\] There are two relevant facts to observe: 1. We have $f = \tilde f \iota_{\hat C}$ where $\iota_{\hat{C}}(c) \da (0, c) \in \hat C$ is inclusion into the second direct summand, and in particular \[ f_n(c) = \hat{f}_n \iota_{\hat C, n}(c) = \hat{f}_n(0, c) .\] \begin{tikzcd} {} \\ {(0, c) \in C_{n-1} \oplus C_{n}} & {} \\ {} & {\hat{C_n}} \\ \\ \\ c & {C_n} &&& {D_n} && {\hat{f}_n(0, c) = f_n(c)} \arrow["{f_n}"', from=6-2, to=6-5] \arrow["{\hat{f}_n}", from=3-2, to=6-5] \arrow["{\iota_{\hat{C}, n}}", from=6-2, to=3-2] \arrow[curve={height=24pt}, dashed, from=6-1, to=6-7] \arrow[dashed, maps to, from=6-1, to=2-1] \arrow[curve={height=-24pt}, dashed, maps to, from=2-1, to=6-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) 2. Since $\hat f$ is a chain map, we have for each $n$ \[ \bd_n^D \hat f_n(x, y) = \hat f_{n+1} \hat{\bd}_n(x, y) && \text{as maps } \hat{C}_n \to D_n .\] We now proceed to compute at the level of elements that $s$ defines a splitting: \[ \bd_{n+1}^D s_n(c) + s_{n-1} \bd_n^D(c) &\da \bd_{n+1}^D \hat{f}_{n+1}(c, 0) + \hat{f}_n \bd_n^C (c, 0) \\ \\ &= \bd_{n+1}^D \hat{f}_{n+1}(c, 0) + {\color{red} \hat{f}_n (\bd_n^C(c), 0) } \\ \\ &= {\color{red} \hat{f_n} \hat{\bd}_{n+1} } (c, 0) + \hat{f}_n (\bd_n^C(c), 0) && \text{by (2)} \\ \\ &= \hat{f}_n \qty{ \begin{bmatrix} -\bd_{n}^C & 0 \\ -\one_{n}^C & \bd_{n+1}^C \end{bmatrix} \begin{bmatrix} c \\ 0 \end{bmatrix} } + \hat{f}_n (\bd_n^C(c), 0) \\ \\ &= \hat{f}_n \qty{ \begin{bmatrix} -\bd_n^C(c) \\ c \end{bmatrix} } + \hat{f}_n (\bd_n^C(c), 0) \\ \\ &= \hat f_n (- \bd_n^C(c), c) + \hat{f}_n( \bd_n^C(c), 0) \\ &= \hat f_n (- \bd_n^C(c) + \bd_n^C(c), c) && \text{since $f_n$ is an $\rmod$ morphism} \\ &= \hat{f}_n(0, c) \\ &= f_n(c) && \text{by (1)} .\] ::: :::{.problem title="Extra"} \envlist a. Show that free implies projective. b. Show that $\Hom_R(M, \wait)$ is left-exact. c. Show that $P$ is projective if and only if $\Hom_R(P, \wait)$ is exact. ::: :::{.solution} \envlist :::{.proof title="of (a)"} Suppose $M$ is free, so we have a set $S$ and an injection $S \injects M$ such that every map in $\hom_{\Set}(S, Y)$ for $Y\in \rmod$ lifts to a unique map in $\hom_{\rmod}(M, Y)$. Suppose further that we have the following situation; we seek to construct a lift $\tilde h: M\to A$: \begin{tikzcd} && {s_i} \\ && S \\ \\ && M \\ \\ A && B && 0 \\ {\alpha_i} && {\beta_i} \arrow["g", from=4-3, to=6-3] \arrow["f"', two heads, from=6-1, to=6-3] \arrow[from=6-3, to=6-5] \arrow["\iota", hook, from=2-3, to=4-3] \arrow["h"{description}, color={rgb,255:red,92;green,214;blue,92}, curve={height=18pt}, from=2-3, to=6-1] \arrow["{\exists ! \tilde h}"{description}, curve={height=12pt}, dashed, from=4-3, to=6-1] \arrow["\in"{marking}, draw=none, from=1-3, to=2-3] \arrow["\in"{marking}, draw=none, from=7-3, to=6-3] \arrow["\in"{marking}, draw=none, from=7-1, to=6-1] \arrow[shift left=5, color={rgb,255:red,92;green,214;blue,92}, curve={height=-30pt}, dotted, maps to, from=1-3, to=7-3] \arrow[color={rgb,255:red,92;green,214;blue,92}, dotted, maps to, from=7-3, to=7-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) This lift exists by first considering $s_i \in S$ and noting that since $\beta_i \da \iota g(s_i) \in B$ and $f$ is surjective, there exist *some* elements \( \alpha_i \) in $A$ such that $f(\alpha_i) = \beta_i$ for each $i$. So define the map \[ h:S &\to A \\ s_i &\mapsto \alpha_i .\] By the universal property of free modules, this lifts to a map $\tilde h: M \to A$, so $M$ is projective. ::: :::{.proof title="of (b)"} Suppose we have a SES \[ 0 \to A \mapsvia{f} B \mapsvia{g} C \to 0 .\] The claim is that $\Hom_R(M, \wait)$ yields an exact sequence \[ 0 \to \Hom_R(M, A) \mapsvia{f_*} \Hom_R(M, B) \mapsvia{g_*} \Hom_R(M, C) \to \cdots ,\] where $f_*(\alpha) = f \circ \alpha$ for \( \alpha:M\to A \) and similarly $g_*(\beta) = g \circ \beta$ for \( \beta:M\to B \). To show that this is exact, it suffices to show three things: 1. $\ker f_* = 0$, 2. $\im f_* \subseteq \ker g_*$, and 3. $\ker g_* \subseteq \im f_*$. Proceeding with each part: 1. By definition, if $\beta: M\to B$ satisfies $\beta \in \ker f_*$, and thus $f\beta = 0$. Since the original sequence was exact, $f$ is injective, thus a monomorphism, thus satisfies the left-cancellation property. So we can immediately conclude that $\beta = 0$. 2. Let $\beta: M\to B$ be in $\im f_*$, so there exists some \( \alpha: M\to A \) with \( \beta = f \alpha \). We want to show that \( \beta\in \ker g_* \), so we can apply $g_*$ to obtain $g_*(\beta) = g_*(f \alpha) \da g f \alpha$. But by exactness of the first sequence, $gf = 0$, so $gfa = 0$. 3. Let \( \beta' \in \ker g_*\), so $g \beta' = 0$. In order to show \( \beta' \in \im f_* \), we want to construct some \( \alpha: M\to A \) such that \( \beta' = f_*( \alpha) \da f \alpha \). Considering $m\in M$, we know \( g \beta'(m) =0 \) and thus \( \beta'(m) \in \ker g = \im f \) by exactness of the first sequence. So there exists some $a_m\in A$ with $f(a_m) = \beta'(m)$, and we can define a map \[ \alpha: M\to A \\ m & \mapsto a_m .\] By construction, we then have \( f \alpha(m) = f(a_m) \da \beta'(m) \) for every $m\in M$, so $f \alpha = \beta'$. ::: :::{.proof title="of (c)"} Assume the same setup as (b). $\implies$: Suppose $P$ is projective, so it satisfies the following universal property: \begin{tikzcd} && P \\ \\ B && C && 0 \arrow["\gamma", from=1-3, to=3-3] \arrow["g"', two heads, from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow["{\exists !\tilde \gamma}"', dashed, from=1-3, to=3-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJCIl0sWzIsMiwiQyJdLFsyLDAsIlAiXSxbNCwyLCIwIl0sWzIsMSwiXFxnYW1tYSJdLFswLDEsImciLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMSwzXSxbMiwwLCJcXGV4aXN0cyAhXFx0aWxkZSBcXGdhbW1hIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d) Using the results of (b), it suffices to check exactness at $\Hom_R(P, C)$ in the following sequence: \[ 0 \mapsvia{} \Hom_R(P, A) \mapsvia{f_*} \Hom_R(P, B) \mapsvia{g_*} \Hom_R(P, C) \to 0 ,\] or equivalently that $g_*$ is surjective. Using that $B \mapsvia{g} C\to 0$ is exact if and only if $g$ is surjective, the universal property above means that every $\gamma \in \Hom_R(P, C)$ lifts to a map $\tilde \gamma \in \Hom_R(P, B)$ where $g\tilde \gamma = \gamma$. Since $g_*(\tilde \gamma) \da g \tilde \gamma$, this precisely means that $\gamma \in \im g_*$. --- $\impliedby$: Reversing the above argument, if $\Hom_R(P, \wait)$ is exact, then every $P \mapsvia{\gamma} C$ has a preimage under $g_*$, which is precisely a lift $P \mapsvia{\tilde \gamma} B$. So $P$ satisfies the universal property of projective modules. ::: :::