# Long Exact Sequence on Homology in an Abelian Category

Let $\mathcal A$ be an abelian category and let $0 \xrightarrow{} X \xrightarrow f Y \xrightarrow g Z \xrightarrow{} 0$ be an exact sequence of chain complexes in $\mathcal A$. I am using the convention that the differentials $d$ go up in degree. By the snake lemma there is a long exact sequence $$\cdots \longrightarrow H^{n} (X) \longrightarrow H^{n}(Y) \longrightarrow H^{n}(Z) \xrightarrow{\ \delta \ } H^{n+1} (X) \longrightarrow \cdots$$ on homology.

Now let $f \colon X \longrightarrow Y$ be any morphism of chain complexes. We define a complex $c(f)$ by $$c(f)^n = X^{n+1} \oplus Y^n \quad \quad d_{c(f)}^n=\left( \begin{array}{cc} -d_X^{n+1} & 0 \\ f^{n+1} & d_Y^n \end{array}\right)$$ and another complex $X[1]$ by $$X[1]^n = X^{n+1} \quad \quad d_{X[1]}^n = - d_X^{n+1}.$$

It is easy to show that $$0 \longrightarrow Y \xrightarrow{ p(f) } c(f) \xrightarrow{q(f)} X[1] \longrightarrow 0$$ is exact where $p(f)^n = \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$ and $q(f)^n = \left( 1 \quad 0 \right)$. Now we can apply the above to this sequence get a long exact sequence on homology, part of which is the connecting morphism $$H^n (X[1]) = H^{n+1}(X) \xrightarrow{\ \delta\ } H^{n+1}(Y)$$ Now apparently this morphism is just $H^{n+1}(f)$. My question is why is this true? The definition of $\delta$ in proofs of the snake lemma that I have seen is very abstract involving pullbacks and pushouts and kernels and cokernels etc. and I am unsure how to show that it must be the same morphism as $H^{n+1}(f)$.

This is covered on the bottom of page 9 / top of page 10 in http://www.scielo.org.ar/pdf/ruma/v48n3/v48n3a01.pdf but I do not understand the quick proof or how the use of elements translates into the language of abelian categories.

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You can use the Freydâ€“Mitchell embedding theorem to justify the use of elements, if you like, or you could use the device Mac Lane uses in Ch. VIII of CWM. You should definitely try to work out the details using the elementary proof of the snake lemma in, say, $\textbf{Ab}$. – Zhen Lin Feb 23 '13 at 22:02
OK ill check out CWM. From what I understand the embedding theorem holds for small abelian categories - does it hold in general? – Paul Slevin Feb 24 '13 at 0:27
You can either take a larger universe, or judiciously choose a small abelian subcategory. (See this question.) – Zhen Lin Feb 24 '13 at 0:45
You can note that the sequence $Y \xrightarrow{p} c(f) \xrightarrow{q} X[1]$ is split exact in each degree. Mimicking the existence proof of the connecting morphism using elements, choose the section $\begin{bmatrix} 1 \\ 0\end{bmatrix}$ of $q^{n}$ and the retraction $\begin{bmatrix} 0 & 1\end{bmatrix}$ of $p^{n+1}$ and note that $f = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} -d & 0 \cr f & d\end{bmatrix} \begin{bmatrix} 1 \cr 0\end{bmatrix}$. Then verify that this is a chain map $X[1] \to Y[1]$ whose homotopy class does not depend on the choice of section and retraction. – Martin Feb 24 '13 at 1:01
You can turn this into a proof with some more work. Here's a heuristic: If you think about how the long exact sequence is constructed using elements, you take a cocycle $x$ in $X^{n+1}$ and take a pre-image $z$ in $c(f)$ so that $q(z) = x$. Then you apply $d_{c(f)}$ which you can check to be in the image of $p$ since $qd_{c(f)} z = d_Xqz = d_Xx = 0$. The pre-image is a representative of $\delta x$. Then you verify that this is independent of the choices. Doing this with the maps I wrote you see that you get the same $\delta x$ and the computation I gave shows this representative to be $f(x)$. – Martin Feb 25 '13 at 20:33