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So far I understand that a long exact sequence refers to a countably infinite exact sequence, and a short exact sequence is a special type of finite exact sequence, beginning and ending with $0$ and containing precisely three nonzero objects in between.

However, I don't understand how the definition of an exact sequence requires finite exact sequences to begin and end with zero. The condition only seems to describe each arrow individually, so seemingly given any finite exact sequence one should be able to remove the leftmost and/or the rightmost spaces as well as their map into/in from the other spaces, and still have an exact sequence.

For example, given a short exact sequence, don't the central three (nonzero) objects and the two arrows between them still form an exact sequence by themselves? So what is the motivation for always including and considering the map from $0$ on the left and the map into $0$ on the right? It seems like redundant information, in addition to being visually unappealing.

And whenever I see examples of finite exact sequences, even not short ones, they always seem to begin and end with $0$ for some reason. What is the motivation for any of this?

I don't know how to use Tikz, so if any of what I am saying or asking is unclear, I will draw, scan, and post example diagrams.

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2 Answers 2

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This is nothing but a nice way to save words.

The defining feature of exact sequences $$ (0\longrightarrow )E_1 \overset{\varphi_1}{\longrightarrow}E_2 \overset{\varphi_2}{\longrightarrow}\dots \overset{\varphi_{n-1}}{\longrightarrow}E_n(\overset{\varphi_n}{\longrightarrow} 0)$$

is the property $\ker\varphi_k=\operatorname{im}\varphi_{k-1}$. If you want to have an exact sequence that starts with an injective map and ends with a surjective one, then obviously $\ker\varphi_1=0$ and $\operatorname{im}\varphi_{n-1}=E_n$. This is nicely encapsulated in the $0$'s at both ends.

If you want to omit the zeros, you may well do so. But then you should say that the first map has no kernel, and the last map is surjective.

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    $\begingroup$ OK, so we always want $\varphi_1$ to be injective and $\varphi_{n-1}$ to be surjective? Because I don't see why that matters in most cases. $\endgroup$ Nov 7, 2016 at 15:30
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    $\begingroup$ @William In the cases where it does not matter then you're free to omit them, obviously. But in many, many cases it does matter. For example, a "short exact sequence $C'_* \to C_* \to C''_*$ does not induce a long exact sequence in homology, but a short exact sequence $0 \to C'_* \to C_* \to C''_* \to 0$ does. This is just one example out of many. Roughly speaking, from a homological algebraic PoV, given an exact sequence $A \to B \to C$ you morally want to consider the homology $\ker(A \to B) \to 0 \to C/B$... Whereas an exact sequence has trivial homology. $\endgroup$ Nov 7, 2016 at 16:21
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    $\begingroup$ It matters because whenever we have a map $A \to B$ and it's not injective, the first thing we want to understand is what the kernel is. And if it's not surjective, we want to understand what the cokernel is. So having a kernel or cokernel of 0 means that we're "done" in that direction (inclusions and surjections are the simplest type of maps). $\endgroup$
    – Ted
    Nov 7, 2016 at 16:25
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There are other situations which give families of exact sequences not quite of the usual form: a fibration of groupoids $p: E \to B$ , see Topology and Groupoids (T&G) Chapter 7, gives a family of exact sequences of the form $$ F_x(x) \xrightarrow{i'} E(x) \xrightarrow{p'}B(y) \xrightarrow{\partial} \pi_0(F_x) \xrightarrow{i_*} \pi_0(E) \xrightarrow{p_*} \pi_0(B).$$ Here $x$ is an object of $E$; $F_x$ is the fibre $p^{-1}(y)$ where $y=p(x)$ .

To say $p$ is a fibration of groupoids is to say that if $x$ is an object of $E$ and $g$ is an arrow of $B$ starting at $p(x)$ then there is an arrow $h$ in $ E$ starting at $x$ such that $p(h)=g$.

The details, and of course the exactness properties, are in that Chapter of T&G, and the idea appeared originally in this 1970 paper. In particular, if $f: Y \to X$ is a fibration of spaces then the induced map of fundamental groupoids is a fibration of groupoids.The case when $f$ is a covering map is also interesting, since then $\pi_1(f)$ is a covering morphism of groupoids. So groupoids give useful algebraic models of some topological phenomena.

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