# Friday, January 29 ## Mapping Cones :::{.remark} Given $f:B\to C$ we defined $\cone(f)_n \da B_{n-1} \oplus C_n$, which fits into a SES \[ 0 \to C \to \cone(f) \mapsvia{\delta} B[-1] \to 0 \] and thus yields a LES in cohomology. \begin{tikzcd} \cdots && {H_{n+1}(\cone(f))} && {H_n(B)} \\ \\ {H_n(C)} && {H_n(\cone(f))} && {H_{n-1}(B)} \\ \\ \cdots \arrow["\delta = f_*", from=1-5, to=3-1, in=180, out=360] \arrow["\delta", from=3-5, to=5-1, in=180, out=360] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=1-3, to=1-5] \arrow[from=1-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMiwwLCJIX3tuKzF9KFxcY29uZShmKSkiXSxbNCwwLCJIX24oQikiXSxbMCwyLCJIX24oQykiXSxbMiwyLCJIX24oXFxjb25lKGYpKSJdLFs0LDIsIkhfe24tMX0oQikiXSxbMCwwLCJcXGNkb3RzIl0sWzAsNCwiXFxjZG90cyJdLFsxLDIsIlxcZGVsdGEiXSxbMiwzXSxbMyw0XSxbMCwxXSxbNSwwXV0=) ::: :::{.corollary title="?"} $f:B\to C$ is a quasi-isomorphism if and only if $\cone(f)$ is exact. ::: :::{.proof title="?"} In the LES, all of the maps $f_*$ are isomorphisms, which forces $H_n(\cone(f)) = 0$ for all $n$. ::: :::{.remark} So we can convert statements about quasi-isomorphisms of complexes into exactness of a single complex. ::: > We'll skip the rest, e.g. mapping cylinders which aren't used until the section on triangulated categories. > We'll also skip the section on \( \delta\dash \)functors, which is a slightly abstract language. ## Ch. 2: Derived Functors :::{.remark} Setup: fix $M\in \rmod$, where $R$ is a ring with unit. Note that by an upcoming exercise, $\Hom_{R}(M, \wait): \modr \to \Ab$ is a *left-exact* functor, but not in general right-exact: given a SES \[ 0\to A \mapsvia{f} B \mapsvia{g} C\to 0 && \in \Ch(\modr) ,\] there is an exact sequence: \begin{tikzcd} 0 && {\Hom_R(M, A)} && {\Hom_R(M, B)} && {\Hom_R(M, C)} \arrow["{f_* = f\circ(\wait)}", from=1-3, to=1-5] \arrow["{g_* = g\circ(\wait)}", from=1-5, to=1-7] \arrow[from=1-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJcXEhvbV9SKE0sIEEpIl0sWzQsMCwiXFxIb21fUihNLCBBKSJdLFs2LDAsIlxcSG9tX1IoTSwgQSkiXSxbMCwwLCIwIl0sWzAsMSwiZl8qID0gZlxcY2lyYyhcXHdhaXQpIl0sWzEsMiwiZ18qID0gZ1xcY2lyYyhcXHdhaXQpIl0sWzMsMF1d) However, this is not generally surjective: not every $M\to C$ is given by composition with a morphism $M\to B$ (*lifting*). To create a LES here, one could use the cokernel construction, but we'd like to do this functorially by defining a sequence functors $F^n$ that extend this on on the right to form a LES: \begin{tikzcd} 0 && {\Hom_R(M, A)} && {\Hom_R(M, B)} && {\Hom_R(M, C)} \\ \\ && {F^1(A)} && {F^1(B)} && {F^1(C)} \\ \\ && {F^2(A)} && \cdots \arrow["{f_* = f\circ(\wait)}", from=1-3, to=1-5] \arrow["{g_* = g\circ(\wait)}", from=1-5, to=1-7] \arrow[from=1-1, to=1-3] \arrow[from=1-7, to=3-3, out=360, in=180] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=5-3, in=180, out=360] \arrow[from=5-3, to=5-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMiwwLCJcXEhvbV9SKE0sIEEpIl0sWzQsMCwiXFxIb21fUihNLCBBKSJdLFs2LDAsIlxcSG9tX1IoTSwgQSkiXSxbMCwwLCIwIl0sWzIsMiwiRl4xKEEpIl0sWzQsMiwiRl4xKEIpIl0sWzYsMiwiRl4xKEMpIl0sWzIsNCwiRl4yKEEpIl0sWzQsNCwiXFxjZG90cyJdLFswLDEsImZfKiA9IGZcXGNpcmMoXFx3YWl0KSJdLFsxLDIsImdfKiA9IGdcXGNpcmMoXFx3YWl0KSJdLFszLDBdLFsyLDRdLFs0LDVdLFs1LDZdLFs2LDddLFs3LDhdXQ==) It turns out such functors exist and are denoted $F^n(\wait) \da \Ext_R^n(M, \wait)$: \begin{tikzcd} 0 && {\Hom_R(M, A)} && {\Hom_R(M, B)} && {\Hom_R(M, C)} \\ \\ && {\Ext_R^1(A)} && {\Ext_R^1(B)} && {\Ext_R^1(C)} \\ \\ && {\Ext_R^2(A)} && \cdots \arrow["{f_* = f\circ(\wait)}", from=1-3, to=1-5] \arrow["{g_* = g\circ(\wait)}", from=1-5, to=1-7] \arrow[from=1-1, to=1-3] \arrow[from=1-7, to=3-3, in=180, out=360] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=5-3, in=180, out=360] \arrow[from=5-3, to=5-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMiwwLCJcXEhvbV9SKE0sIEEpIl0sWzQsMCwiXFxIb21fUihNLCBBKSJdLFs2LDAsIlxcSG9tX1IoTSwgQSkiXSxbMCwwLCIwIl0sWzIsMiwiXFxFeHRfUl4xKEEpIl0sWzQsMiwiXFxFeHRfUl4xKEIpIl0sWzYsMiwiXFxFeHRfUl4xKEMpIl0sWzIsNCwiXFxFeHRfUl4yKEEpIl0sWzQsNCwiXFxjZG90cyJdLFswLDEsImZfKiA9IGZcXGNpcmMoXFx3YWl0KSJdLFsxLDIsImdfKiA9IGdcXGNpcmMoXFx3YWl0KSJdLFszLDBdLFsyLDRdLFs0LDVdLFs1LDZdLFs2LDddLFs3LDhdXQ==) By convention, we set $\Ext_R^0(\wait) \da \Hom_R(M, \wait)$. This is an example of a general construction: **right-derived functors** of $\Hom_R(M, \wait)$. More generally, if \( \mathcal{A} \) is an abelian category (with a certain additional property) and $F: \mathcal{A} \to \mathcal{B}$ is a left-exact functor (where \( \mathcal{B} \) is another abelian category) then we can define right-derived functors $R^n F: \mathcal{A} \to \mathcal{B}$. These send SESs in \( \mathcal{A} \) to LESs in \( \mathcal{B} \): \begin{tikzcd} 0 && A && B && C && 0 \\ \\ 0 && FA && FB && FC \\ \\ && {R^1FA} && {R^1 FB} && {R^1 FC} \\ \\ && \cdots \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=5-3, in=180, out=360] \arrow[from=5-3, to=5-5] \arrow[from=5-5, to=5-7] \arrow[from=5-7, to=7-3, in=180, out=360] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTMsWzAsMCwiMCJdLFsyLDAsIkEiXSxbNCwwLCJCIl0sWzYsMCwiQyJdLFs4LDAsIjAiXSxbMCwyLCIwIl0sWzIsMiwiRkEiXSxbNCwyLCJGQiJdLFs2LDIsIkZDIl0sWzIsNCwiUl4xRkEiXSxbNCw0LCJSXjEgRkIiXSxbNiw0LCJSXjEgRkMiXSxbMiw2LCJcXGNkb3RzIl0sWzAsMV0sWzEsMl0sWzIsM10sWzMsNF0sWzUsNl0sWzYsN10sWzcsOF0sWzgsOV0sWzksMTBdLFsxMCwxMV0sWzExLDEyXV0=) Similarly, if $F$ is *right-exact* instead, there are left-derived functors $L^n F$ which form a LES ending with 0 at the right: \begin{tikzcd} 0 && A && B && C && 0 \\ &&&&&& \cdots \\ && LFA && LFB && LFC \\ \\ && FA && FB && FC && 0 \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=5-3, to=3-7, in=360, out=180] \arrow[from=5-3, to=5-5] \arrow[from=5-5, to=5-7] \arrow[from=5-7, to=5-9] \arrow[from=3-3, to=2-7, in=360, out=180] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTMsWzAsMCwiMCJdLFsyLDAsIkEiXSxbNCwwLCJCIl0sWzYsMCwiQyJdLFs4LDAsIjAiXSxbMiwyLCJMRkEiXSxbNCwyLCJMRkIiXSxbNiwyLCJMRkMiXSxbMiw0LCJGQSJdLFs0LDQsIkZCIl0sWzYsNCwiRkMiXSxbOCw0LCIwIl0sWzYsMSwiXFxjZG90cyJdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs1LDZdLFs2LDddLFs3LDhdLFs4LDldLFs5LDEwXSxbMTAsMTFdLFs1LDEyXV0=) ::: ## 2.2: Projective Resolutions :::{.definition title="Projective Modules"} Let \( \mathcal{A} = \rmod \), then \( P \in \rmod \) satisfies the following universal property: \begin{tikzcd} && P \\ \\ B && C && 0 \arrow["g", from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow["{\exists \beta}"', dashed, from=1-3, to=3-1] \arrow["\gamma", from=1-3, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJCIl0sWzIsMiwiQyJdLFs0LDIsIjAiXSxbMiwwLCJQIl0sWzAsMSwiZyJdLFsxLDJdLFszLDAsIlxcZXhpc3RzIFxcYmV0YSIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDEsIlxcZ2FtbWEiXV0=) ::: :::{.remark} Free modules are projective. Let $F = R^X$ be the free module on the set $X$. Then consider $\gamma(x)\in C$, by surjectivity these can be pulled back to some elements in $B$: \begin{tikzcd} && X \\ \\ && F \\ \\ B && C && 0 \\ {\exists b\in g^{-1}(\gamma(x)) \da \beta(x)} && {\gamma(x)} \arrow["g", from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \arrow["{\exists \beta}"', dashed, from=3-3, to=5-1] \arrow["\gamma", from=3-3, to=5-3] \arrow["{\iota_X}", hook, from=1-3, to=3-3] \arrow["{\exists \tilde \beta}"', dotted, from=1-3, to=5-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCw0LCJCIl0sWzIsNCwiQyJdLFs0LDQsIjAiXSxbMiwyLCJGIl0sWzIsNSwiXFxnYW1tYSh4KSJdLFswLDUsIlxcZXhpc3RzIGJcXGluIGdeey0xfShcXGdhbW1hKHgpKSBcXGRhIFxcYmV0YSh4KSJdLFsyLDAsIlgiXSxbMCwxLCJnIl0sWzEsMl0sWzMsMCwiXFxleGlzdHMgXFxiZXRhIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzMsMSwiXFxnYW1tYSJdLFs2LDMsIlxcaW90YV9YIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNiwwLCJcXGV4aXN0cyBcXHRpbGRlIFxcYmV0YSIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRvdHRlZCJ9fX1dXQ==) This follows from the universal property of free modules: \begin{tikzcd} &&&& {\exists F(X)} \\ \\ \\ X &&&& M & {\in \rmod} \arrow["{f\in \Hom_\Set(X, M)}", from=4-1, to=4-5] \arrow["{\exists g\in \Hom_\Set(X, F(X))}", from=4-1, to=1-5] \arrow["{\exists ! f' \in \Hom_R(F(X), X)}", dashed, from=1-5, to=4-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwzLCJYIl0sWzQsMywiTSJdLFs0LDAsIlxcZXhpc3RzIEYoWCkiXSxbNSwzLCJcXGluIFxccm1vZCJdLFswLDEsImZcXGluIFxcSG9tX1xcU2V0KFgsIE0pIl0sWzAsMiwiXFxleGlzdHMgZ1xcaW4gXFxIb21fXFxTZXQoWCwgRihYKSkiXSxbMiwxLCJcXGV4aXN0cyAhIGYnIFxcaW4gXFxIb21fUihGKFgpLCBYKSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) ::: :::{.proposition title="Projective if and only if summand of free (for modules)"} An \(R\dash\)module is projective if and only if it is a direct summand of a free module. ::: :::{.exercise title="?"} Prove the $\impliedby$ direction! ::: :::{.proof title="?"} $\implies$: Assume $P$ is projective, and let $F(P)$ be the free \(R\dash\)module on the underlying set of $P$. We can start with this diagram: \begin{tikzcd} &&&& {F(P)} \\ \\ \\ P &&&& P \arrow["{\id_P}", from=4-1, to=4-5] \arrow[from=4-1, to=1-5] \arrow[dashed, from=1-5, to=4-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwzLCJQIl0sWzQsMywiUCJdLFs0LDAsIkYoUCkiXSxbMCwxLCJcXGlkX1AiXSxbMCwyXSxbMiwxLCIiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) And rearranging, we get \begin{tikzcd} &&&&&& P \\ \\ 0 && {\ker \pi} && {F(P)} && P && 0 \arrow["\pi", two heads, from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow["\id", from=1-7, to=3-7] \arrow["{\exists \iota}"{description}, from=1-7, to=3-5] \arrow[from=3-1, to=3-3] \arrow[hook, from=3-3, to=3-5] \arrow["\iota", curve={height=-18pt}, dashed, from=3-7, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbNCwyLCJGKFApIl0sWzYsMiwiUCJdLFs4LDIsIjAiXSxbNiwwLCJQIl0sWzIsMiwiXFxrZXIgXFxwaSJdLFswLDIsIjAiXSxbMCwxLCJcXHBpIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzEsMl0sWzMsMSwiXFxpZCJdLFszLDAsIlxcZXhpc3RzIFxcaW90YSIsMV0sWzUsNF0sWzQsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwwLCJcXGlvdGEiLDAseyJjdXJ2ZSI6LTMsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) Since \( \pi \circ \iota \), the SES splits and this $F(P) \cong P \oplus \ker \pi$, making $P$ a direct summand of a free module. ::: :::{.example title="?"} Not every projective module is free. Let $R = R_1 \cross R_2$ a direct product of unital rings. Then $P \da R_1 \cross \ts{0}$ and $P' \da \ts{0} \cross R_2$ are \(R\dash\)modules that are submodules of $R$. They're projective since $R$ is free over itself as an \(R\dash\)module, and their direct sum is $R$. However they can not be free, since e.g. $P$ has a nonzero annihilator: taking $(0, 1)\in R$, we have $(0, 1) \cdot P = \ts{(0, 0)} = 0_R$. No free module has a nonzero annihilator, since ix $0\neq r\in R$ then $rR \neq 0$ since $r 1_R\in r R$, which implies that $r \qty{ \bigoplus R } \neq 0$. ::: :::{.example title="?"} Taking $R = \ZZ/6\ZZ = \ZZ/2\ZZ \oplus \ZZ/3\ZZ$ admits projective \(R\dash\)modules which are not free. ::: :::{.example title="?"} Let $F$ be a field, define the ring $R \da \Mat(n \cross n, F)$ with $n\geq 2$, and set $V = F^n$ thought of as column vectors. This is left \(R\dash\)module, and decomposes as $R = \bigoplus _{i=1}^n V$ corresponding to the columns of $R$, using that $AB = [Ab_1, \cdots, Ab_n]$. Then $V$ is a projective \(R\dash\)module as a direct summand of a free module, but it is not free. We have vector spaces, so we can consider dimensions: $\dim_F R = n^2$ and $\dim_F V = n$, so $V$ can't be a free \(R\dash\)module since this would force $\dim_F V = kn^2$ for some $k$. ::: :::{.example title="?"} How many projective modules are there in a given category? Let \( \mathcal{C}\da \Ab^\fin \) be the category of *finite* abelian groups, where we take the full subcategory of the category of all abelian groups. This is an abelian category, although it is not closed under *infinite* direct sums or products, which has no projective objects. :::{.proof title="?"} Over a PID, every submodule of a free module is free, and so we have free $\iff$ projective in this case. So equivalently, we can show there are no free $\ZZ\dash$modules, which is true because $\ZZ$ is infinite, and any such module would have to contain a copy of $\ZZ$. ::: ::: :::{.remark} The definition of projective objects extends to any abelian category, not just \(R\dash\)modules. :::