# Abstract Nonsense

## The Long Exact Sequence

Point of Post: In this post we construct the, ever-powerful, long-exact sequence of homology objects induced by a short exact sequence of chain complexes.

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Motivation

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We shall discuss one of the most useful aspects of homology–it’s relation with long-exact sequences of chain complexes. Throughout all of mathematics being able to put an object $X$ in the middle of a short exact sequence $0\to Y\to X\to Z\to 0$ where $Y,Z$ are known allows one to glean information about $X$ from information about $Y,Z$. Why should chain complexes be any different? Namely, if it’s computationally advantageous to try to put a module or a group in the middle of a short exact sequence, it stands to reason that putting a chain complex in such a sequence would be equally fruitful. While this is all just $R\text{-}\mathbf{Mod}$ intuition it actually turns out to be quite correct in general. Namely, we shall see that the ability to form a short exact sequence $0\to \mathbf{C}'\to\mathbf{C}\to\mathbf{C}''\to 0$ shall tell us that there is a fairly substantial connection between the homology objects of $\mathbf{C},\mathbf{C}'$, and $\mathbf{C}''$. This shall be very useful computationally but it shall also enable us to prove some pretty theoretical things as well–in particular it shall prove to be a very powerful tool in relating certain levels of the derived functors we are going to create. It also has huge uses in topology in as basic ideas as the long exact sequence of relative homology groups.

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Hopefully the above motivates you to care about what I am about to talk about. Unfortunately, this is perhaps famous for being one of the most annoying, messy, and just nonsensey part of homological algebra. It is just infeasible to prove this result directly in general categories and so we are going to be forced to prove it for $R\text{-}\mathbf{Mod}$ and appeal to the metatheorem to give us the general result.

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The Snake Lemma and the Long Exact Sequence

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Before we actually construct the connection between the homology groups of a short exact sequence of chain complexes in some module category we are going to prove a certain lemma. I actually separate this from the rest of the post because it is useful in its own right. This is the infamous snake lemma. The reason why the snake lemma is so famous (beside’s it’s downright, damnable annoyingess) is that it is (perhaps) one of the highest levels of mathematics displayed in a major motion picture. It was in 1980 that the researchers for the movie It’s My Turn decided that the snake lemma was the perfect thing to have their love-confused mathematician main character teach the snake lemma as an indication of her mathematical prowess. The result was this. While not the most precise mathematical scene imaginable (jumping from the snake lemma to group cohomology?) it definitely stands against any other example churned out by the silver screen. Indeed, while Mr. Cooperman is the classic, parodied, nerdy math grad student his concerns are valid. The part of the proof that he is questioning is, undoubtedly, the worst part of proving the snake lemma. Moreover, after his little tiff he (I think) asks the main character about her progress with her “$2$-fusion group” which is a real thing–although, it’s entirely feasible that he said Fuchsian group.

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Regardless, it’s time to actually state and prove this damnable lemma, and we can’t get a real proof from our love-crossed mathematician (although, try telling this to Weibel):

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Theorem(The Snake Lemma): Let $A,B,C,D,E,F$ be $R$-modules and

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$\begin{array}{ccccccccc} & & A & \xrightarrow{\alpha} & B & \xrightarrow{\beta} & C & \to & 0\\ & & _{f}\big\downarrow & & _{g}\big\downarrow & & _{h}\big\downarrow & & \\ 0 & \to & D & \xrightarrow{\gamma} & E & \xrightarrow{\delta} & F & & \end{array}$

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a commutative diagram of $R$-maps where the rows are exact. Then, there exists an  exact sequence

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$\ker f\to\ker g\to \ker h\to\text{coker }f\to\text{coker }g\to\text{coker }h$

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where the maps $\ker f\to\ker g$ and $\text{coker }g\to\text{coker h}$ are mono and epi if $\alpha$ and $\delta$ are respectively.

Proof: The only map that is kind of scary is the map $\ker h\to\text{coker }f$. Indeed, let us first verify that everything between the other modules works nicely. For example let’s verify that we can take the maps $\ker f\to\ker g$ and $\ker g\to\ker h$ to just be $\alpha$ and $\beta$ respectively. Indeed, we must merely check that $\alpha(\ker f)\subseteq\ker g$ and similarly for $\beta$. But, if $x\in\ker f$ we know that $g(\alpha(x))=\gamma(f(x))=0$ and so $\alpha(x)\in\ker g$–the case for $\beta$ is similar. Similarly, let’s show that we can take the maps $\text{coker }f\to\text{coker }g$ and $\text{coker }g\to\text{coker }h$ to be $x+\text{im }f\mapsto \gamma(x)+\text{im }g$ and $x+\text{im }g\mapsto \delta(x)+\text{im }h$ respectively. To see that this is well-defined we merely note that if $\gamma(f(x))=g(\alpha(x))$ so that if $x$ and $y$ differ by an element of $\text{im }f$ their images will differ by an element of $\text{im }g$ in $E$ and so everything is well-defined. The same thing works for the map $\text{coker }g\to\text{coker }h$.

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Ok, so we at least know what our maps are at all points of this exact sequence we are supposed to create except the map $\ker h\to\text{coker }f$. This map is annoying and a little contrived, but here we go. We start with $x\in\ker h$, since $\beta$ is surjective there exists $b\in B$ with $\beta(b)=\ker h$. Now, note that $0=h(\beta(b))=\delta(g(b))$ and thus we know that $g(b)\in\ker\delta=\text{im }\gamma$ so there exists a unique (because of injectivity) $d\in D$ such that $\gamma(d)=g(b)$. We then define the map $\partial:\ker h\to\text{coker }f$ by $\partial(x)=d+\text{im }f$. Of course, we have to verify that this map is independent of our initial choice of $b$. Indeed, suppose for a second that $\beta(b')=x$ and let $d'$ be the element of $D$ such that $\gamma(d')=g(b')$.  Note then that $b-b'\in\ker\beta$ and so there exists $q\in A$ such that $\alpha(q)=b-b'$. Thus, $g(b-b')=g(\alpha(q))=\gamma(f(q))$. Thus, we see that $g(b-b')=g(b)-g(b')=\gamma(d)-\gamma(d')=\gamma(d-d')$ and thus we see by injectivity that $f(q)=d-d'$ and thus we see that $d+\text{im }f=d'+\text{im }f$. This proves that $\partial$ is well-defined.

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We now prove that $\partial$ is a homomorphism. To see this we merely let $x,x'\in\ker h$ and $r,s\in R$. Let $b,b'\in B$ be such that $\beta(b)=x$ and $\beta(b')=x'$. Note then that

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$\gamma(r\partial(x)+s\partial(x'))=r\gamma(\partial(x))+s\gamma(\partial(x'))=rg(b)+sg(b')=g(rb+sb')$

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Since $\beta(rb+sb')=rx+sx'$ this allows us to conclude that $\partial(rx+sx')=r\partial(x)+s\partial(x')$ and thus $\partial$ is a homomorphism.

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Thus, we have all our maps and we just have to check exactness. All of these are routine with, perhaps, the exception of exactness at $\ker h$, and so this is the only calculation that we show. To show that $\text{im }\beta_{\mid\ker g}\subseteq\ker\partial$ we let $x\in B$ and merely note that if we can find $a$ such that $\gamma(a)=g(x)$ then $\partial(\beta(x))=a$ since $x$ is an element of $B$ such that $\beta(x)=\beta(x)$ (funny, right?)–but $g(x)=0$ and so we may take $a=0$ and so $\partial(\beta(x))=0$. To show the reverse inclusion we let $x\in\ker\partial$.  Since $\partial(x)=0$ there exists $b\in B$ such that $\beta(b)=x$ and $g(b)=\gamma(f(q))$ since $\partial(x)\in\text{im }f$ (by definition of being zero in $\text{coker }f$). But, we see then that $g(b)=\gamma(f(q))=g(\alpha(q))$ and so $b-\alpha(q)\in\ker g$ and $\beta(b-\alpha(q))=x-\beta(\alpha(q))=x$. Form where it follows that $x\in\beta_{\mid\ker g}$. $\blacksquare$

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As we have mentioned by the metatheorem this is now actually proven in general abelian categories.  We use this then to prove the following theorem

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Theorem: Let $\mathscr{A}$ be an abelian category and suppose we have the following diagram with exact rows

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$\begin{array}{ccccccccc}0 & \to & A_n' & \xrightarrow{\alpha} & A_n & \xrightarrow{\beta} & A_n'' & \to & 0\\ & & ^{d_n'}\big\downarrow & & ^{d_n}\big\downarrow & & \big\downarrow^{d_n''} & & \\ 0 & \to & A_{n-1}' & \xrightarrow{\gamma} & A_{n-1} & \xrightarrow{\delta} & A_{n-1}'' & \to & 0\end{array}\quad\mathbf{(1)}$

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(where the diagram extends infinitely above and below where indicated [for all $n\in\mathbb{Z}$]) Then then we a diagram of the form

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$\begin{array}{ccccccccc} & & \text{coker}(d_{n+1}') & \to & \text{coker }(d_{n+1}) & \to & \text{coker}(d_{n+1}'') & \to & 0\\ & & ^{d_n'}\big\downarrow & & \big\downarrow^{d_n} & & \big\downarrow^{d_n''} & & \\ 0 & \to & \ker d_{n-1}' & \to & \ker d_{n-1} & \to & \ker d_{n-1}'' & & \end{array}\quad\mathbf{(2)}$

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with exact rows.

Proof: This is obvious in the categories $R\text{-}\mathbf{Mod}$ for a ring $R$, and so by the metaprinciple we are done. $\blacksquare$

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Now we can finally construct the long exact sequence from an exact chain of chain complexes:

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Theorem: Let $\mathscr{A}$ be an abelian category and $\mathbf{0}\to\mathbf{C}'\xrightarrow{\{f_n\}}\mathbf{C}\xrightarrow{\{g_n\}}\mathbf{C}''\to\mathbf{0}$ be an exact sequence in $\mathbf{Ch}(\mathscr{A})$. Then, there exists a long exact sequence of homology objects

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\begin{aligned}\cdots\to H_{n+1}(\mathbf{C}')\xrightarrow{f_{n+1}}H_{n+1}(\mathbf{C})\xrightarrow{g_{n+1}}H(\mathbf{C}'')\\ & \quad \xrightarrow{c_{n+1}}H_n(\mathbf{C}')\xrightarrow{f_n}H_n(\mathbf{C})\xrightarrow{g_n} H(\mathbf{C}'')\to\cdots\end{aligned}

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Proof: It’s clear that we only need to construct the maps $c_n$ that make everything work, since these are the only possible bad points. To construct such maps we note that from the exact sequence of chain complexes we get a sequence like in $\mathbf{(1)}$ (with $C$‘s instead of $A$‘s and $\partial$‘s instead of $d$‘s) and thus we get a diagram like in $\mathbf{(2)}$ which when we apply the snake lemma gives us an exact sequence like we want. $\blacksquare$.

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For modules the maps $H_n(\mathbf{C}'')\to H_{n-1}(\mathbf{C}')$ are given explicitly by $z+\text{im }\partial_n''\mapsto f_{n-1}^{-1}(d_n(p_n^{-1}(z)))+\text{im }\partial_n'$ where the “inverse” denotes any element in the preimage–it’s independent of choice.

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References:

[1] Weibel, Charles A. An Introduction to Homological Algebra. Cambridge [England: Cambridge UP, 1994. Print.

[2] Schapira, Pierre. “Categories and Homological Algebra.” Web. <http://people.math.jussieu.fr/~schapira/lectnotes/HomAl.pdf&gt;.

[3] Rotman, Joseph. An Introduction to Homological Algebra. Dordrecht: Springer, 2008. Print.