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What is the motivation for requiring that the square of a differential be $0$ for a complex, aside from enabling us to speak of the homology of a complex?

Other homological notions like chain maps, homotopic maps, homotopy equivalences seem to be meaningful without any restriction on the differential (of course, no longer do homotopic maps induce isomorphisms on Homology, for Homology no longer is meaningful).

If we ignore any connections to Homology, is there some other moral reason to want that the differential square to $0$?

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    $\begingroup$ The word "moral" is totally inapplicable here, but something like "intuitive" or "heuristic" would do. $\endgroup$ – rschwieb May 20 at 20:32
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    $\begingroup$ Like you said, it enables us to speak of the homology of a complex. $\endgroup$ – anomaly May 29 at 18:10
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    $\begingroup$ @rschwieb "morally" and "moral" are actually often used to mean something similar to "intuitive" (but not exactly the same thing -althoug it's still a vague, unprecise notion), in some places at least. OP : Cartan famously said "if I could only understand the beautiful consequences following from the concise proposition $d^2=0$". An unsatisfactory answer would be : homology theory was built when many examples of this "$d^2=0$" phenomenon were found, like for instance the differential of the de Rham complex (unsatisfactory because it's probably not historically accurate) $\endgroup$ – Max May 29 at 18:13
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    $\begingroup$ From the other end, why a lot of commonly occurring complexes do have $d^2 = 0$: it often comes from the fact that an exact complex (for example a resolution) has $d^2 = 0$, and the fact that applying any functor preserves the condition that $d^2 = 0$ (while not that many functors preserve exactness). $\endgroup$ – Daniel Schepler May 29 at 18:18
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    $\begingroup$ There are certainly situations where one has $d^n=0$, and then a variant on homology (or several variants, depending on $i$) can be defined as $\textrm{ker} d^i/\textrm{im} d^{n-i}$. $\endgroup$ – John Palmieri May 29 at 23:21
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The main motivation for wanting $d \circ d = 0$ is because it arises naturally in so many applications. As others have already said in comments, exact sequences are common and important, and you often get chain complexes by applying functors to exact sequences. At that point, homology measures how far the new sequence is from being exact: often a very interesting question.

More generally, people certainly study diagrams of the form $$ X_1 \to X_2 \to X_3 \to \cdots $$ with no condition on the maps, and then you can examine maps between such diagrams — the analog of chain maps. This happens all the time in algebra; for example, $X_i$ might be a submodule of $X_{i+1}$ for each $i$, giving a filtration of $\bigcup X_i$.

In the case when $d^2 \neq 0$, although I suppose the analog of "chain homotopy" could be defined, it is not clear what use it would be. What can you deduce if $f$ and $g$ are "chain homotopic" in this more general sense? What does it tell you in the case where each $X_i \to X_{i+1}$ is injective? Is surjective? When considering honest chain complexes, the definition of chain homotopy is motivated by the definition of homotopy in topology, and indeed, homotopic maps between topological spaces induce chain homotopic maps on their singular chain complexes: there is a mechanism for producing chain homotopies, at least in this one situation. Is there any mechanism that produces this analog of a chain homotopy? Without good motivating examples and/or interesting consequences, it doesn't seem worthwhile (to me, at least) to devote too much energy to it.

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  • $\begingroup$ “It arises in so many applications”. Could you list two or three? The only place I’ve ever seen it in particular is with differential forms. Also, when you refer to some relationship with exact sequences, it is a complete mystery to me what the connection is, so I imagine it might be the case to others too. Could you explain that briefly in the solution? $\endgroup$ – rschwieb May 30 at 23:19
  • $\begingroup$ @rschwieb: in algebraic topology, there are several ways to produce a chain complex from a suitable topological space: the singular, cellular, and simplicial chain complexes. Also in any exact sequence, you have $d^2=0$. Then if you apply a functor, say $- \otimes_{\mathbb{Z}} G$ for your favorite abelian group $G$, you will get a sequence which may no longer be exact but has $d^2=0$. Read about homological algebra and group cohomology for two large areas which are replete with chain complexes and exact sequences. $\endgroup$ – John Palmieri May 31 at 1:50
  • $\begingroup$ @rschwieb: If you like more examples coming from analysis, take a look at Cousin problems which first appear as questions in complex analysis but they are best understood in terms of sheaf cohomology (which also provides some tools for solving the problems). $\endgroup$ – Moishe Kohan May 31 at 3:50
  • $\begingroup$ @JohnPalmieri That still flew far, far over my head, so I took a look at the wiki article. I was surprised to see that $d^2=0$ is apparently a shorthand for something I had not expected, namely pairwise relationships between the arrows in a sequence. Is there some precise sense in which this is really an operator that squares to zero, or have I just been mislead by a convention? $\endgroup$ – rschwieb Jun 3 at 17:14
  • $\begingroup$ I don't understand your question. Which parts flew over your head? Regarding $d^2 = 0$, can you clarify your question? It's in the title of the question here, and the setup should be a chain complex in which each map is called $d$, so the composite of two consecutive ones is $d \circ d := d^2$. $\endgroup$ – John Palmieri Jun 4 at 17:17

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