9
$\begingroup$

Previously I asked How did we know to invent homological algebra?, because I was under the misapprehension that concrete examples of long exact sequences had been a major motivation for developing homological algebra. I've now learned to my surprise that they were invented in the 1940's, long after other parts of homological algebra were better-established.

In the comments there, @Qiaochu Yuan suggested I ask the variant question How should we have known to invent homological algebra?; taking his suggestion, I'm asking it now.

Of course this question is open to a considerable amount of interpretation.

$\endgroup$
3
$\begingroup$

Consider the extension problem in group theory: we say that $K$ is an extension of $G$ by $A$ is there is a surjective homomorphism $\phi: K \to G$ with $\ker(\phi) = A$. How many extension of $G$ are there by $A$ up to some good notion of equivalence?

Suppose we have such an extension in which all involved groups are abelian. Consider that $\hom(G,A)$ naturally injects into $\hom(G,K)$ so we can form the quotient $\hom(G,K)/\hom(G,A)$. In addition, the extension $\phi$ induces a map $\phi_*:\hom(G,K) \to \hom(G,G)$ by composition. It is clear that kernel of this map includes the image of $\hom(G,A)$ in $\hom(G,K)$, because $\ker(\phi) = A$. So is $\hom(G,G) \simeq \hom(G,K)/\hom(G,A)$?

No: just consider $G = \mathbb{Z}_2$, $K = \mathbb{Z}$ and $A = 2\mathbb{Z}$.

We could phrase the extension problem differently: given (abelian) groups $G$ and $A$, an extension of $G$ by $A$ is a group $K$ admitting an injection $i: A \to K$ and a surjection $\phi: K \to A$ such that $\ker(\phi) = i(A)$. Consider that $\hom(K,A)$ naturally surjects onto $\hom(A,A)$ by restriction. Form the quotient $\hom(K,A)/\hom(A,A)$. In addition, the map $\phi$ induces a map $\phi_*: \hom(K,A) \to \hom(G,A)$ by composition. It is clear that the kernel of this map includes $\hom(A,A)$ because $\ker(\phi) = A$. So is $\hom(G,A) \simeq \hom(K,A)/\hom(A,A)$?

No: just consider the example (reframed to this extension problem).

What gives? More specifically, what are the relevant cokernels?

The first difficulty arises from lifting: $G \simeq K/A$, but there are self-maps of $G$ which don't lift to maps from $G$ to $K$. The second difficulty arises from extension: there are maps from $G$ to $A$ which don't extend to maps from $K$ to $A$.

As groups are $\mathbb{Z}$-modules, you could try modules over your other favorite ring. What if you try to change ring via tensor product?


As for topology, your original question has an answer from Sammy Black which uses no exact sequences. Again, the issue is that a particular construction does not play nicely with quotients.

This is my best attempt at an answer. The lesson to me is that it's a good idea to think through these homological constructions without reference to an exact sequence. If you view the sequences as organizational tools, then "the functor $\mathcal{F}$ is not exact" just means that "the functor $\mathcal{F}$ does not work nicely with out organizational scheme." But relative homology and group extension are both quite natural without any fancy diagrams.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.