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Update: Qiaochu Yuan points out in the comments that the title of the question is misleading, as homological algebra did not begin with long exact sequences as I'd thought.

(Original question follows.)

I want to understand how anyone knew that the long exact sequence in homology was something to go looking for. It seems like every time I push it back a step I just end up with more questions, though...

I know that the long exact sequence in homology comes from short exact sequences of chain complexes. The motivating example I've generally seen for this is something like this: Let $B$ be a (simplicial, CW, etc.) complex and let $A$ be a subcomplex of $A$. The inclusion induces a map of chain complexes $C_\bullet A \to C_\bullet B$, and so we can get a short exact sequence $0 \to C_\bullet A \to C_\bullet B \to C_\bullet B / C_\bullet A \to 0$. If we write $H_\bullet(B, A) := H_\bullet(C_\bullet B / C_\bullet A)$ then we get induced maps $H_n(A) \to H_n(B) \to H_n(B, A)$.

Ignore for the moment that it seems to be a pretty far stretch to get from there to the idea that a long exact sequence might exist in homology. So far as I can tell $H_n(B, a)$ is not a natural object to study from a geometric perspective, and the only purpose of defining it is that it (tautologically) gives us a short exact sequence.

But then how do we know that short exact sequences are something we want to study? The only motivation I know of for wanting to look at a short exact sequence is that it can lead to a long exact sequence in homology, but how do you discover that fact without knowing to start by looking at short exact sequences?

Of course even once we've started looking at short exact sequences, the existence of a connecting map seems far from obvious.

Can anyone give some insight as to how people ever came up with this stuff?

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  • $\begingroup$ I don't know any homological algebra so maybe this isn't what you're looking for. But I know looking at short exact sequences is sometimes a useful way to look at a normal subgroup of a group and the associated quotient group: $N\rightarrow G\rightarrow G/N$. $\endgroup$ – Thoth Jun 10 '13 at 19:10
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    $\begingroup$ The obvious pre-topology idea that leads to homology theory is the Euler characteristic. A truly remarkable thing, that this numeric value is calculating something about a space, independent of how that space is split up into polygons and polytopes. The Euler characteristic is some kind of invariant, but of what? Attempting to answer that question, and to generalize it, leads to homology. $\endgroup$ – Thomas Andrews Jun 10 '13 at 19:22
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    $\begingroup$ @ThomasAndrews: I'm on board with why we want to study homology groups. What I don't understand is how we realized that we could study them using what's now known as "homological algebra." Does the Euler characteristic explain why we'd anticipate that such a thing as the long exact sequence in homology would exist? $\endgroup$ – Daniel McLaury Jun 10 '13 at 19:28
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    $\begingroup$ In my opinion the idea that a connecting map could exist between the homology groups is not so obvious if one deals with a short exact sequence of chain complexes of arbitrary abelian groups, but in topology it pretty much suggests itself, because it has very easy discription: You take the class of a relative cycle $\alpha$ in $X$, then just send it to class of its boundary in $A$. It seems also very plausible that this connecting homomorphism has image equal to the kernel of $i_*$, for if $[\alpha]\in H(A)$ is zero in $H(X)$ simply if it is the boundary of some $\beta$. $\endgroup$ – Stefan Hamcke Jun 10 '13 at 19:28
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    $\begingroup$ I think the motivation is to somehow relate homology of $X$, the subspace $A$, and the quotient space $X/A$. A cycle in $X$ is a chain which has no boundary, and if we collapse $A$ to a point, a chain with boundary in $A$ has its boundary shrunk to a single point, which is "almost nothing" (actually it is empty if $n$ is odd, and a constant $n-1$-simplex if $n$ is even). So in order to get an idea of the cycles in the quotient space, one had to look at the relative cycles in $(X,A)$. $\endgroup$ – Stefan Hamcke Jun 10 '13 at 19:41
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In my opinion, the relative singular homology group $H_n(B, A)$ is naturally a geometric object. As Stefan H. says in the comments, the idea of considering the boundary of a relative cycle in $(A, B)$ as a cycle of one dimension less in $B$ is quite natural.

But why is this connected to the short exact sequence? A quotient of chain groups (the relative group) ought to be related to a quotient of the underlying spaces.

In the relative group $C_n(B, A)$, the condition for a chain to be a cycle is relaxed from the boundary being $0$ to the boundary being a chain in the subspace $A$. Under mild assumptions on the pair $(A, B)^\dagger$, the quotient map $B \to B/A$ induces an isomorphism $$ H_n(B, A) \overset{\sim}{\longrightarrow} \tilde{H}_n(B/A). $$ When the subspace $A$ is quotiented to a point, the boundary of a chain in $B$ maps to that point in $H_n(B/A)$, or $0$ in the reduced group.


$^\dagger$ $A$ is closed and is a deformation retract of a neighborhood $A'$ in $B$

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    $\begingroup$ I think $A$ should be closed and a deformation retract of a neighborhood. $\endgroup$ – Stefan Hamcke Jun 10 '13 at 20:07
  • $\begingroup$ Yes, @StefanH. Thanks. $\endgroup$ – Sammy Black Jun 10 '13 at 20:09
  • $\begingroup$ To expand on this a bit, apparently the sequence $H^n(A) \to H_n(B) \to H_n(B / A) \to H_{n-1}(A) \to \cdots$ was considered first, and then the relative homology group was introduced retroactively because it made things work out more smoothly. (See my comment on the other answer.) $\endgroup$ – Daniel McLaury Jun 11 '13 at 19:51
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Have you read Weibel's History of Homological Algebra?

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  • $\begingroup$ Weibel has exactly two sentences on the topic, namely that the concept first appears in an abstract of a 1941 talk by Hurewicz. That abstract reads: $\endgroup$ – Daniel McLaury Jun 11 '13 at 16:36
  • $\begingroup$ [...] Consider "natural homomorphisms" of $H^n(A) \to H^n(B) \to H^{n+1}(A - B) \text{[sic]} \to H^{n+1}(A) \to H^{n+1}(B) \to H^{n+1}(A - B)$. It can be shown that the kernel of each of these homomorphisms is the image of the preceding homomorphism. This statement contains Kolmogorov's generalization of Alexander's duality theorem and has many applications. [...] $\endgroup$ – Daniel McLaury Jun 11 '13 at 16:39
  • $\begingroup$ I'd have preferred a bit more detail, but I guess this at least validates what Stefan H. is saying. $\endgroup$ – Daniel McLaury Jun 11 '13 at 16:42
  • $\begingroup$ @Daniel: I think the bigger point is that the answer to the title question and the body question diverge. The historical roots of homological algebra are not in chain complexes and exact sequences; they were in relatively concrete topological and then algebraic phenomena that people cared about, and they gradually realized that they could use certain machinery to understand that phenomena. But the historical roots are in the phenomena, not the machinery. $\endgroup$ – Qiaochu Yuan Jun 11 '13 at 18:55
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    $\begingroup$ @Daniel: if you want to understand the machinery more, I think the question should be edited to "how should we have known to invent homological algebra?" which I think is an interesting and very different question. I was going to write a series of blog posts about this at some point but lost interest... $\endgroup$ – Qiaochu Yuan Jun 11 '13 at 18:56
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As I know, "ancient" topologists had studied Betti numbers $\beta_n(A)$ (the ranks of abelian groups $H_n(A)$) and then they noticed the connection between sequences $\{\beta_n(A)\}$ and $\{\beta_n(B)\}$.

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