# Lecture 5 (Monday, January 25) ## LES Associated to a SES :::{.theorem title="Every SES of chain complexes induces a LES in homology"} For every SES of chain complexes, there is a long exact sequence in homology. ::: :::{.proof title="?"} Suppose we have a SES of chain complexes \[ 0 \to A \mapsvia{f} B \mapsvia{g} C \to 0 ,\] which means that for every $n$ there is a SES of \(R\dash\)modules. Recall the diagram for the snake lemma, involving kernels across the top and cokernels across the bottom. Applying the snake lemma, by hypothesis $\coker g = 0$ and $\ker f = 0$. There is a SES \[ A_n / d A_{n+1} \to B_n / d B_{n+1} \to C_n / d C_{n+1} \to 0 \] Using the fact that $B_n \subseteq Z_n$, we can use the 1st and 2nd isomorphism theorems to produce \begin{tikzcd} & {H_n(A)} & {H_n(B)} & {H_n(C)} \\ \\ & {A_n/d A_{n+1}} & {B / d B_{n+1}} & {C/d C_{n+1}} && 0 \\ \\ 0 & {Z_{n-1}(A)} & {Z_{n-1}(B)} & {Z_{n-1}(C)} \\ \\ & {\substack{\coker d_n \\ = Z_{n-1}(A)/d A_n \\ = H_{n-1}(A)} } & {H_{n-1}(B)} & {H_{n-1}(C)} \arrow[from=3-4, to=3-6] \arrow["f", from=3-2, to=3-3] \arrow["g", from=3-3, to=3-4] \arrow["{d_n}"', from=3-2, to=5-2] \arrow["{d_n}"', from=3-3, to=5-3] \arrow["{d_n}"', from=3-4, to=5-4] \arrow[from=5-2, to=7-2] \arrow["{f_*}"', from=7-2, to=7-3] \arrow["{g_*}"', from=7-3, to=7-4] \arrow[from=5-3, to=7-3] \arrow[from=5-4, to=7-4] \arrow["{f_*}"{description}, from=1-2, to=1-3] \arrow["{g_*}"{description}, from=1-3, to=1-4] \arrow[from=1-2, to=3-2] \arrow[from=1-3, to=3-3] \arrow[from=1-4, to=3-4] \arrow[from=5-1, to=5-2] \arrow[from=5-2, to=5-3] \arrow[from=5-3, to=5-4] \arrow[dotted, from=1-4, to=7-2, in=180, out=360, "{\exists \delta}"] \end{tikzcd} > [Link to diagram](https://q.uiver.app/?q=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) This yields an exact sequence relating $H_n$ to $H_{n-1}$, and these can all be spliced together. - $\ker(A_n / d A_{n-1} \to Z_{n-1}(A) = Z_n(A) / d A_{n+1} \da H_n(A)$ using the 2nd isomorphism theorem ::: :::{.remark} Note that $d$ is *natural*, which means the following: there is a category \( \mathcal{S} \) whose objects are SESs of chain complexes and whose maps are chain maps: % https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsxLDAsIkEiXSxbMiwwLCJCIl0sWzMsMCwiQyJdLFs0LDAsIjAiXSxbMCwyLCIwIl0sWzEsMiwiQSciXSxbMiwyLCJCJyJdLFszLDIsIkMnIl0sWzQsMiwiMCJdLFsxLDZdLFsyLDddLFszLDhdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs1LDZdLFs2LDddLFs3LDhdLFs4LDldXQ== \begin{tikzcd} 0 & A & B & C & 0 \\ \\ 0 & {A'} & {B'} & {C'} & 0 \arrow[from=1-2, to=3-2] \arrow[from=1-3, to=3-3] \arrow[from=1-4, to=3-4] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \end{tikzcd} There is another full subcategory \( \mathcal{L} \) of $\Ch$ whose objects are LESs of objects in the original abelian category, i.e. exact chain complexes. The claim is that the LES construction in the theorem defines a functor \( \mathcal{S}\to \mathcal{L} \). We've seen how this maps objects, so what is the map on morphisms? Given a morphism as in the above diagram, there is an induced morphism: \begin{tikzcd} \cdots & {H_n (A)} & {H_n(B)} & {H_n(C)} & {H_{n-1}(A)} & \cdots \\ \\ \cdots & {H_n(A')} & {H_n(B')} & {H_n(C')} & {H_{n-1}(A')} & \cdots \arrow["\bd", from=1-4, to=1-5] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow["\bd", from=3-4, to=3-5] \arrow["{H_n(u_A)}", from=1-2, to=3-2] \arrow["{H_n(u_B)}", from=1-3, to=3-3] \arrow["{H_n(u_C)}", from=1-4, to=3-4] \arrow["{H_{n-1}(u_A)}", from=1-5, to=3-5] \arrow[from=3-5, to=3-6] \arrow[from=1-5, to=1-6] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTIsWzAsMCwiXFxjZG90cyJdLFsxLDAsIkhfbiAoQSkiXSxbMiwwLCJIX24oQikiXSxbMywwLCJIX24oQykiXSxbNCwwLCJIX3tuLTF9KEEpIl0sWzAsMiwiXFxjZG90cyJdLFsxLDIsIkhfbihBJykiXSxbMiwyLCJIX24oQicpIl0sWzMsMiwiSF9uKEMnKSJdLFs0LDIsIkhfe24tMX0oQScpIl0sWzUsMCwiXFxjZG90cyJdLFs1LDIsIlxcY2RvdHMiXSxbMyw0LCJcXGJkIl0sWzAsMV0sWzEsMl0sWzIsM10sWzUsNl0sWzYsN10sWzcsOF0sWzgsOSwiXFxiZCJdLFsxLDYsIkhfbih1X0EpIl0sWzIsNywiSF9uKHVfQikiXSxbMyw4LCJIX24odV9DKSJdLFs0LDksIkhfe24tMX0odV9BKSJdLFs5LDExXSxbNCwxMF1d) The first two squares commute, and *naturality* means that the third square commutes as well. ::: :::{.exercise title="?"} Check the details! ::: :::{.remark} It is sometimes useful to explicitly know how to compute snake lemma boundary elements. See the book for a recipe for computing $\bd(\xi)$: - Lift $\xi$ to a cycle $c\in Z_n(C) \subseteq C_n$. - Pull $c$ back to a preimage $b\in B_n$ by surjectivity. - Apply the differential to get $d(b)\in Z_{n-1}(B)$, using that images are contained in kernels. - Since this is in kernel of the outgoing map, it's in the kernel of the incoming map and thus there exists an $a\in Z_{n-1}(A)$ such that $f(a) = db$ - So set \( \delta(\xi) \da [a] \in H_{n-1}(A) \). ::: :::{.remark} Why is naturality useful? Suppose $H_n(B) = 0$, you get isomorphisms, and this allows inductive arguments up the LES. The LES in homology is sometimes abbreviated as an **exact triangle**: % https://q.uiver.app/?q=WzAsMyxbMSwwLCJIXyooQSkiXSxbMiwyLCJIXyooQikiXSxbMCwyLCJIXyooQykiXSxbMCwxLCJmIl0sWzEsMiwiZyJdLFsyLDAsIlxcYmQiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJzcXVpZ2dseSJ9fX1dXQ== \begin{tikzcd} & {H_*(A)} \\ \\ {H_*(C)} && {H_*(B)} \arrow["f", from=1-2, to=3-3] \arrow["g", from=3-3, to=3-1] \arrow["\bd", squiggly, from=3-1, to=1-2] \end{tikzcd} Here $\bd:H_*(C) \to H_*(A)[1]$ shifts degrees. Note that this motivates the idea of **triangulated categories**, which is important in modern research. See Weibel Ch.10, and exercise 1.4.5 for how to construct these as quotients of $\Ch$. ::: ## 1.4: Chain Homotopies :::{.remark} Assume for now that we're in the situation of \(R\dash\)modules where $R$ is a field, i.e. vector spaces. The main fact/advantage here that is not generally true for \(R\dash\)modules: every subspace has a complement. Since $B_n \subseteq Z_n \subseteq C_n$, we can write $C_n = Z_n \oplus B_n'$ for every $n$, and $Z_n = B_n \oplus H_n$. This notation is suggestive, since $H_n \cong Z_n/B_n$ as a quotient of vector spaces. Substituting, we get $C_n = B_n \oplus H_n \oplus B_n'$. Consider the projection $C_n \to B_n$ by projecting onto the first factor. Identifying $B_n \da \im(C_{n+1} \to C_n) \cong C_{n+1}/Z_{n+1}$ by the 1st isomorphism theorem in the reverse direction. But this image is equal to $B_{n+1}'$, and we can embed this in $C_{n+1}$, so define $s_n: C_n \to C_{n+1}$ as the composition \[ s_n \da ( C_n \mapsvia{\proj} B_n = \im (C_{n+1} \to C_n) \mapsvia{d_{n+1}\inv} C_{n+1}/Z_{n+1} \mapsvia{\cong} B_{n+1}' \injects C_{n+1} .\] ::: :::{.claim title="1"} $d_{n+1} s_n d_{n+1} = d_{n+1}$ are equal as maps. ::: :::{.proof title="?"} \envlist - Check on the first factor $B_{n+1}' \subseteq C_{n+1}$ directly to get $s_n d_{n+1}(x) = d_{n+1}(x)$ for $x\in B_{n+1}'$, and then applying $d_{n+1}$ to both sides is the desired equality. - On the second factor $Z_{n+1}$, both sides give zero since this is exactly the kernel. ::: :::{.claim title="2"} $d_{n+1} s_n + s_{n-1}d_n = \id_{C_n}$ if and only if $H_n = 0$, i.e. the complex $C$ is exact at $C_n$. This map is the sum of taking the two triangle paths in this diagram: % https://q.uiver.app/?q=WzAsNCxbMiwwLCJDX24iXSxbNCwwLCJDX3tuLTF9Il0sWzIsMiwiQ19uIl0sWzAsMiwiQ197bisxfSJdLFsxLDIsInNfe24tMX0iXSxbMCwyLCJcXGlkIiwyXSxbMCwxLCJkX24iXSxbMywyLCJkX3tuKzF9Il0sWzAsMywic19uIl1d \begin{tikzcd} && {C_n} && {C_{n-1}} \\ \\ {C_{n+1}} && {C_n} \arrow["{s_{n-1}}", from=1-5, to=3-3] \arrow["\id"', from=1-3, to=3-3] \arrow["{d_n}", from=1-3, to=1-5] \arrow["{d_{n+1}}", from=3-1, to=3-3] \arrow["{s_n}", from=1-3, to=3-1] \end{tikzcd} ::: :::{.proof title="?"} We again check this on both factors: - Using the first claim, $s_n = 0$ on $B_n'$ and thus $s_{n-1} d_n = \id_{B_n'}$. - On $H_n$, $s_n = 0$ and $d_n = 0$, and so the LHS is $0 = \id_{H_n}$ *if and only if* $H_n = 0$. - On $B_n$, and tracing through the definition of $s_n$ yields $d_{n+1} s_n(x) = x$ and this yields $\id_{B_n}$. ::: Next time: summary of decompositions, start general section on chain homotopies.