# Monday, February 22 ## Colimits and Adjoints :::{.proposition title="Characterizations of cocomplete categories"} Assume \( \cat{A} \) is abelian so we have cokernels for maps. TFAE: 1. \( \bigoplus A_i \) exists in \( \cat{A} \) for every set \( \ts{A_i} \) of objects in \( \cat{A} \). 2. \( \cat{A} \) is cocomplete, i.e. \( \colim_{i\in I}A_i \) exists for every functor \( \cat{I} \to \cat{A} \) with \( \cat{I} \) small. ::: :::{.proof title="?"} Note that (1) is a special case of (2), so it suffices to show \( 1\implies 2 \). Given a functor \( A: \cat{I} \to \cat{A} \) and let \( f: \bigoplus _{\alpha i\to j} A_i \to \bigoplus_{i\in \cat{I}} A_i \) where $i,j \in \cat{I}$. \begin{tikzcd} i && j && {\cat{I}} \\ \\ {A_i} && {A_j} && {\cat{A}} \arrow["\alpha", from=1-1, to=1-3] \arrow["{\alpha_*}", from=3-1, to=3-3] \arrow["A"{description}, squiggly, from=1-3, to=3-3] \arrow["A"{description}, squiggly, from=1-1, to=3-1] \arrow["A", squiggly, from=1-5, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJpIl0sWzIsMCwiaiJdLFswLDIsIkFfaSJdLFsyLDIsIkFfaiJdLFs0LDIsIlxcY2F0e0F9Il0sWzQsMCwiXFxjYXR7SX0iXSxbMCwxLCJcXGFscGhhIl0sWzIsMywiXFxhbHBoYV8qIl0sWzEsMywiQSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6InNxdWlnZ2x5In19fV0sWzAsMiwiQSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6InNxdWlnZ2x5In19fV0sWzUsNCwiQSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6InNxdWlnZ2x5In19fV1d) Then the map \( f( a_{i, \alpha}) = \alpha_*(a_i) - a_i \in A_j - A_i\), so this is \( \alpha_* - \one \). Let \( C\da \coker f \da \bigoplus_{i\in I} A_i / \im(f) \), and we'll denote elements in this quotient with a bar. :::{.claim} $C = \colim_{i\in I} A_i$ with \[ \eta_i: A_i &\to C \\ a_i &\mapsto \bar{a_i} ,\] where we first embed $A_i$ into the direct sum and then take the quotient. ::: :::{.exercise title="?"} Use the universal property of cokernels in \( \cat{A} \). Check that the following diagram commutes: % https://q.uiver.app/?q=WzAsMyxbMCwwLCJBX2kiXSxbMCwyLCJBX2oiXSxbMiwxLCJDIl0sWzAsMiwiXFxldGFfaSJdLFsxLDIsIlxcZXRhX2oiLDJdLFswLDEsIlxcYWxwaGFfKiIsMl1d \begin{tikzcd} {A_i} \\ && C \\ {A_j} \arrow["{\eta_i}", from=1-1, to=2-3] \arrow["{\eta_j}"', from=3-1, to=2-3] \arrow["{\alpha_*}"', from=1-1, to=3-1] \end{tikzcd} This essentially follows from the fact that \( \bar{ \alpha_*(a_i)} = \bar{a_i} \). ::: ::: :::{.remark} $\modr$ satisfies (1), since direct sums of \(R\dash\)modules still have an \(R\dash\)module structure. Thus \( \modr \) is cocomplete. ::: :::{.definition title="Limits"} The **limit** of a functor \( A:\cat{I} \to \cat{A}\) is the colimit of the dual functor \( A\op : I\op \to \cat{A}\op \). ::: :::{.remark} Note that this amounts to reversing arrows in the conditions of a colimit. Many of the results for colimits go through with arrows reversed. Examples: kernels, direct products. If $I$ is a poset, then limits are referred to as **inverse limits**, using $\inverselim_{i\in I} A_i$. ::: :::{.definition title="Complete Categories"} \( \cat{A} \) is **complete** if and only if \( \lim_{i\in I} A_i \) exists whenever \( \cat{I} \) is small and \( A: \cat{I} \to \cat{A} \). ::: :::{.theorem title="The Adjoint-Limit Theorem"} Let \( \adjunction{L}{R}{ \cat{A} }{ \cat{ B} } \) be an adjoint pair, where now \( \cat{A}, \cat{B} \) are now arbitrary categories (not necessarily abelian). Then - The **left adjoint** $L$ preserves **colimits** (direct sums, cokernels, etc). I.e. if \( A: \cat{I} \to \cat{A} \) has a colimit, then so does \( (L \circ A ): \cat{I}\to \cat{B} \), and \( L( \colim A_i) = \colim (LA_i) \). - The **right adjoint** $R$ preserves **limits** (direct products, kernels, etc). ::: :::{.proof title="?"} Not given in the book! See MacLane's *Categories for the Working Mathematician*. ::: :::{.remark} Recall left adjoints are right-exact and have left-derived functors. ::: :::{.corollary title="?"} If \( \cat{A} \) is a cocomplete abelian category with enough projectives and \( \adjunction{F}{G}{ \cat{A} } { \cat{B} } \). Then for every set-indexed collection of objects \( \ts{ A_i } \), \[ (L_* F)\qty{ \bigoplus_{i \in I } A_i } = \bigoplus _{i\in I} L_* F(A_i) ) ,\] so left-derived functors commute with direct sums. ::: :::{.proof title="?"} Let $P_i$ be the projective resolution of $A_i$, so $P_i \to A_i$, then \( \bigoplus P_i \to \bigoplus A_i \) is a projective resolution, and by definition \[ (L_* F) \qty{ \bigoplus A_i} &= H_*\qty{ F\qty{ \bigoplus P_i } } \\ &= H_* \qty{ \bigoplus FP_i } \quad \text{by the theorem} \\ &\cong \bigoplus H_*( FP_i) \text{homology commutes with $\oplus \in \Ch(\cat{A})$} \\ &= \bigoplus _i L_* F(A_i) .\] ::: :::{.corollary title="?"} For $A_i \in \rmod, B\in \modr$, \[ \Tor_*^R\qty{ \bigoplus_{i\in I} A_i, B } \cong \bigoplus_{i\in I} \Tor_*^R(A_i, B) .\] ::: :::{.proof title="of corollary"} \[ \Tor_*^R(\wait, B) = L_*F, && F \da (\wait \tensor_R B) ,\] and $F$ is a left-adjoint by the tensor-hom adjunction. ::: :::{.remark} One can also show directly from the definition that \[ \Tor_*^R(A, \bigoplus_{i\in I} B_i) \cong \bigoplus_{i\in I} \Tor_*^R(A, B_i) .\] This uses the fact that \( P \tensor_R (\bigoplus_{i\in I} B_i) \cong \bigoplus_{i\in I} (P \tensor B_i) \). ::: :::{.remark} We'll skip the rest of this section, we (hopefully) won't need filtered colimits. ::: ## Balancing Tor and Ext :::{.remark} Idea: their derived functors with either variable fixed will essentially be the same. We'll start by showing that the two left-derived functors of $\wait \tensor_R \wait$ give the same results, and similarly for the two right-derived functors $\Hom_R(\wait, \wait)$. We'll use double complexes! ::: ### Tensor Product Complexes :::{.remark} Suppose we have two chain complexes \( (P)_R \in \Ch( \modr ), {}_R(Q) \in \Ch(\rmod) \). Then there is a double complex where $i, j$ indexes rows and columns: \( P \tensor_R Q = \ts{ P_i \tensor_R Q_j }_{i, j} \), the **tensor product double complex** of $P$ and $Q$. We use the sign trick from 1.2.5: - $d^h \da d^P \tensor \one$ - $d^v \da (-1)^i 1 \tensor d^Q$ Taking the direct sum totalization \( \Tor^{ \oplus }(P \tensor_R Q ) \) is the **total tensor product chain complex** of $P$ and $Q$. Note that this has a single differential! The big theorem from this section: ::: :::{.theorem title="Tor is balanced"} \[ L_n(A\tensor_R \wait)(B) \cong L_n(\wait \tensor_R B)(A) \da \Tor_n^R(A, B) .\] ::: :::{.remark} Note that this makes the right-hand side notation unambiguous. ::: :::{.proof title="?"} Choose projective resolutions \( P \mapsvia{\eps} A \in \modr \) and \( Q \mapsvia{\eta} B \in \rmod \). We'll form 3 tensor product double complexes. - $P\tensor Q$: A first quadrant double complex, since the projective resolutions have nonnegative indices. - $A\tensor Q$, embedding $A \embeds \Ch( \cat {A} )$ as a complex concentrated in degree 0 (so one column) - $P \tensor B$ (one row). There are several maps of double complexes among these induced by \( \epsilon, \eta \): \begin{tikzcd} && {} \\ && {} & \vdots && \vdots && \vdots \\ & {A \tensor Q_2} && {P_0 \tensor Q_2} && {P_1\tensor Q_2} && {P_2 \tensor Q_2} && \cdots \\ & {A \tensor Q_1} && {P_0 \tensor Q_1} && {P_1\tensor Q_1} && {P_2 \tensor Q_1} && \cdots \\ & {A \tensor Q_0} && {P_0 \tensor Q_0} && {P_1\tensor Q_0} && {P_2 \tensor Q_0} && \cdots \\ {} && {} &&&&&& {} \\ &&& {P_0 \tensor B} && {P_1 \tensor B} && {P_2 \tensor B} \\ && {} \arrow["{\epsilon \tensor 1}", from=3-4, to=3-2] \arrow["{\epsilon \tensor 1}", from=4-4, to=4-2] \arrow["{\epsilon \tensor 1}", from=5-4, to=5-2] \arrow["{1 \tensor \eta}"{description}, from=5-4, to=7-4] \arrow["{1 \tensor \eta}"{description}, from=5-6, to=7-6] \arrow["{1 \tensor \eta}"{description}, from=5-8, to=7-8] \arrow[from=2-4, to=3-4] \arrow["{d^P \tensor 1}"', from=3-6, to=3-4] \arrow[from=2-6, to=3-6] \arrow[from=2-8, to=3-8] \arrow["{d^P \tensor 1}"', from=3-8, to=3-6] \arrow["{d^P \tensor 1}", from=4-8, to=4-6] \arrow["{d^P \tensor 1}", from=5-8, to=5-6] \arrow["{d^P \tensor 1}", from=5-6, to=5-4] \arrow["{d^P \tensor 1}", from=4-6, to=4-4] \arrow["{1\tensor d^Q}", from=3-4, to=4-4] \arrow["{1\tensor d^Q}", from=4-4, to=5-4] \arrow["{1\tensor d^Q}", from=3-6, to=4-6] \arrow["{1\tensor d^Q}", from=4-6, to=5-6] \arrow["{1\tensor d^Q}", from=3-8, to=4-8] \arrow["{1\tensor d^Q}", from=4-8, to=5-8] \arrow[draw={rgb,255:red,214;green,92;blue,92}, dotted, no head, from=6-1, to=6-9] \arrow[draw={rgb,255:red,214;green,92;blue,92}, dotted, no head, from=1-3, to=8-3] \arrow[from=5-10, to=5-8] \arrow[from=4-10, to=4-8] \arrow[from=3-10, to=3-8] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) We'll show there are two maps: \[ A \tensor Q = \Tot(A\tensor Q) \fromvia{\eps\tensor \one} \Tor(P\tensor Q) \mapsvia{1\tensor\eta} \Tor(P\tensor B) = P\tensor B ,\] using that totalizing a one-row or one-column complex is summing along diagonals where each has one term, yielding actual equality of the first and last terms respectively above. Moreover, we'll show these are **quasi-isomorphisms**, and so \[ L_*(A\tensor \wait) \fromvia{\eps \tensor \one} H_*( \Tor (P\tensor Q) ) \mapsvia{\one \tensor \eta} L_*( \wait \tensor B)(A) .\] We'll continue with the proof of this next time. :::