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I am having trouble with finding the best intuition for homology as sometimes I'm able to find for other subjects which is how someone could have come up with it.

Reading the history of homology didn't satisfy me, I was wondering if someone could help me with this. (The other answers on math exchange aren't really what I'm looking for, they somehow already assume we know what the homology groups are).

I would love an answer that is to homology as those short ones (mostly because I can't think of a long one) illustrate the kind of motivation I like best:

  1. Homotopy: It is clearly interesting to see the maps from $S^1$ to our space look like. After some thinking, we realize we can impose a group structure if we fix the base point, but do we really want to? Umm we kind of get stuck otherwise, so let's fix a basepoint to make things simpler, in fact we also see (details left) after some thinking that it doesn't matter which one if the space is connected!

    Ok let's take a simple example like the square, umm well this is already very complicated, but we don't want this, since the square deformation retracts to a point so we want stuff like that to be simple. Okay so we actually want maps from $S^1$ up to homotopy, and our group is still works, and there we have given a motivation to the definition of homotopy group.

2.Discrete finite fourier basis: We hate the regular basis, we want orthonormal stuff!!! We know that if we multiply polynomials the convolution theorem exists (which is how we get fast multiplication), maybe this is true in general? Yes!

Hopefully this made the kind of motivation I'm looking for clear. I'm unable to find one for homology :(.

I understand that we first want higher homotopy groups but they turn out to be hard, so the next thing is it's obviously interesting how maps from $[0,1]^n$ to our space look like, and with a bit of a stretch I can believe we want to linear algebra\group everything, so we consider finite sums of those maps, and from technical reasons we may later find out it's easier to work with maps from the simplex.

However, this still leaves open how we can come up with the boundary maps, or why we would consider the homology groups (I understand that in a weird sense they measure holes- including that the first homology is the abelinzation of the fundemental group), but I would like more convincing how someone could think of this is as the "obvious" next step (Even though historically I'm aware this wasn't the case, it's easier for me to find an explanation of course with retrospect which makes this the obvious next step).

edit: I of course don't require that the answer be a continuation of what I started about homology, a lot of times good examples (if you can show how someone could come from Euler's formula to homology) generalize well to give a good motivation, but I haven't exactly understood how this is the case here.

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    $\begingroup$ Why is the history not satisfactory? That how it actually came about. $\endgroup$ – Rene Schipperus Dec 13 '17 at 21:51
  • $\begingroup$ @ReneSchipperus In this case I don't find it satisfying; sometimes the history is very long and shows how tricky and nontrivial something is, as is it seems in this case, but often I found that with a lot of retrospect we can give a decent explanation of how something is the "obvious" thing you'd want to do when you're trying to do "this", and this is very satisfying. Gower's blog is a good example of this imo, see gowers.wordpress.com/2007/09/15/… , gowers.wordpress.com/2007/09/19/… $\endgroup$ – Andy Dec 13 '17 at 21:58
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    $\begingroup$ If I got it right you asked about the origin of singular homology. Surely it all started with Euler characteristic and then simplicial homology, which is much easier to describe. The boundary maps are about the actual boundary of a simplex! It's just a bit harder to prove that the simplicial homology is an invariant of the space. Don't forget about de Rham cohomology either! After all, it also measures how holes disturb the closed vs exact differential forms (or vector fields in 3D). $\endgroup$ – Jyrki Lahtonen Dec 13 '17 at 22:01
  • $\begingroup$ @JyrkiLahtonen Yes, thanks (I now clarified it in the title). I am willing ot take the Euler characteristic as given, but I still don't see how we progress from it to singular\simplicial homology (If I understand correctly, I'd be actually satisfied with getting motivation for simplicial homology if it's simpler, since I feel like we can come to singular homology once we understand why simplicial homology is "obvious" as first a technical tool to show independece of tringulation and later a nic concept). $\endgroup$ – Andy Dec 13 '17 at 22:05
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    $\begingroup$ See assorteddetails.wordpress.com/2016/04/27/what-is-homology $\endgroup$ – Balarka Sen Dec 13 '17 at 22:28
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I know this answer will diverge somewhat from the topological flavor of the original question - but I thought I'd share some thoughts on how somebody "could have" come up with homological algebra from an algebraic point of view.

So, suppose you have a short exact sequence $0 \to A \to B \to C \to 0$ of, say, $R$-modules; then you have a morphism $f : U \to C$, and you want to determine whether there is a lifting of this morphism to a morphism $\tilde f : U \to B$. (If you like, you can suppose that $R$ is Noetherian, and each module is finitely generated.) Once you realize this is not always possible, but it is possible if $U$ is a free $R$-module, then you might start to investigate how to determine from $U$ what sorts of "obstructions" there are to constructing a lifting map.

So, you would start off by taking some set of generators of $U$, and then try to map each generator $x$ to some preimage of $f(x)$ in $B$. Now, the reason this doesn't always work is that there are also relations in $U$, and each of these relations has to map to zero in $B$. On the other hand, all we really know from being given that $f$ is a well-defined morphism is that they map to zero in $C$; so from the short exact sequence, we can conclude that the images of the relations in $B$ are actually in $A$.

To formalize this, we can let $F^0$ be the free $R$-module generated by our chosen generators of $U$, and then let $F^1$ be the free $R$-module generated by one symbol for each "relation" between these generators. This gives the start of a free resolution of $U$, $F^1 \to F^0 \to U \to 0$. Now, the argument from the previous paragraph gives a morphism in $\operatorname{Hom}(F^1, A)$. However, just because this map might be nonzero for some choice doesn't necessarily mean we're doomed: if we can find a map in $\operatorname{Hom}(F^0, A)$ which composes to give our map in $\operatorname{Hom}(F^1, A)$, then we can subtract that map from our candidate map $F^0 \to B$ to fix it up into a map $F^0 \to B$ which does induce a well-defined map $\tilde f : U \to B$; and because the fix-up map has image contained in $A$, it doesn't affect the images of generators of $U$ in $C$.

Therefore, we can construct an object classifying the obstructions as $\mathscr{O} := \operatorname{Hom}(F^1, A) / \operatorname{im}(\operatorname{Hom}(F^0, A))$, along with an "obstruction class" map $\operatorname{Hom}(U, C) \to \mathscr{O}$. It is not too hard to prove that a map in $\operatorname{Hom}(U, C)$ can be lifted to a map $\operatorname{Hom}(U, B)$ if and only if its obstruction class is zero, giving a one-element extension of the exact sequence you get from left exactness of $\operatorname{Hom}(U, \cdot)$: \begin{equation} 0 \to \operatorname{Hom}(U, A) \to \operatorname{Hom}(U, B) \to \operatorname{Hom}(U, C) \to \mathscr{O}. \end{equation} You might have also noticed that the construction of $\mathscr{O}$ only depends on $A$ and on the free resolution of $U$, so we might give it some name like $\operatorname{Ext}^1(F^\cdot, A)$.

Now, from here, you might wonder if it extends to $\operatorname{Ext}^1(F^\cdot, B)$. As it turns out, using our definition that would not extend the long exact sequence; however, from here, it isn't too much of a stretch to imagine some clever researcher noticing that if you extend the free resolution, and take the subobject $\ker(\operatorname{Hom}(F^1, A) \to \operatorname{Hom}(F^2, A))$ and then do the same quotient, then the sequence does extend to a long exact sequence of six objects. And moreover, that once you do this, the result becomes independent of the choice of free resolution of $U$. From here, the next step of defining $\operatorname{Ext}^i(U, A) := \ker(\operatorname{Hom}(F^i, A) \to \operatorname{Hom}(F^{i+1}, A)) / \operatorname{im}(\operatorname{Hom}(F^{i-1}, A) \to \operatorname{Hom}(F^i, A))$ and getting an infinite long exact sequence isn't much of a leap beyond that.

This reflects a general theme in (co)homology theory: that an $H^1$ or $H_1$ object is often closely related, via the first boundary map, to a class of obstructions to doing some "lifting" or "gluing" construction.

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    $\begingroup$ Hmm, after all the time I spent typing this up, then I notice the title specifically mentions singular homology. Maybe an edit during the time I was writing the answer? Now, I'm wondering whether I should just leave it here; or delete it and possibly copy the contents to another more related question, or create a new question for it if I can't find a suitable existing question. $\endgroup$ – Daniel Schepler Dec 13 '17 at 23:15
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    $\begingroup$ Please leave it :), I am still reading, but I take any information and intuition I can put my hands on (especially when I'm sure that when I'll see this appear in algebra I'll have the same question about homological algebra) $\endgroup$ – Andy Dec 13 '17 at 23:18
  • $\begingroup$ Very pretty! I wasn't even aware people found this interesting just as an algebraic object. Do you think there is a way to relate this to the singular homology? $\endgroup$ – Andy Dec 13 '17 at 23:57
  • $\begingroup$ Well, yes there are some relations at a couple degrees of separation. For example, a generalization of this $\operatorname{Ext}^\cdot$ construction leads to the notion of "left-derived" or "right-derived" functors. Then, a special case of that gives $H^\cdot(X, \mathscr{F})$, "cohomology with coefficients from a sheaf of values". Then, if you can prove the sheaf of local functions from simplices to $G$ is an "acyclic resolution" (i.e. $H^i = 0$ for $i > 0$ and $H^0 \simeq G$), then you can conclude that the singular cohomology is identical to the sheaf cohomology of the constant sheaf $G$. $\endgroup$ – Daniel Schepler Dec 14 '17 at 0:08
  • $\begingroup$ OK, to be more precise, "acyclic resolution" means $H^i(A^j) = 0$ whenever $i > 0$, and the sequence $0 \to G \to A^0 \to A^1 \to \cdots$ is exact. $\endgroup$ – Daniel Schepler Dec 14 '17 at 0:12
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Answering how the boundary operator is a natural thing to consider in homology: Homology (singular, simplicial, and to a lesser extent, cellular) basically asks whether one can, given an $n$-sphere in a space, fill in the sphere to make an $(n+1)$-ball. It's a very natural intrinsic way to look for "holes" in the space.

Rephrasing, we're asking whether something looking like the boundary of an $(n+1)$-ball actually is the boundary of such a ball. How do we tell that something looks like the boundary of a ball? That it is $n$-dimensional, doesn't have boundary itself, and is orientable (which may be taken care of by giving a sign to the boundary operator) is a good start.

That turns out to be more or less enough as well. A thing that fulfills these criteria might not look like the boundary of a sphere (it could be a torus, for instance). But whatever it looks like the boundary of, that thing can be glued together from balls. And because of the sign convention on the boundary operator, in the places where the balls touch, the boundaries from each of them cancel out. So while what we study by the description above isn't necessarily balls and spheres, it can always be built from balls and spheres.

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  • $\begingroup$ Thanks for the answer. So the main thing about boundery is that we want to detect orientable stuff, not just any holes, is that correct? In what way would be able to detect any holes, not just orientable ones? Maybe with the coefficients in $Z_2$? Is this true? If so, can you give an example of a hole we'd find? $\endgroup$ – Andy Dec 13 '17 at 22:13
  • $\begingroup$ @Andy There is nothing here stopping you from detecting unorientable stuff. For instance, $H_1(\Bbb RP^2)=\Bbb Z_2$ is detectable just fine. The only reason I put orientable in there is because you don't expect to be able to "fill in" the inside of something unorientable. If I embed a Klein bottle in a space, it almost doesn't matter what the space looks like, the bottle will never be the boundary of a ball. So we require our potential boundaries to be orientable just to avoid those problems. $\endgroup$ – Arthur Dec 13 '17 at 22:26
  • $\begingroup$ Then I still don't understand the signs when we take boundry (I see on small examples like a circle that it cancels out), but not why in general why we take them signed (and not coefficients in $Z_2$). $\endgroup$ – Andy Dec 13 '17 at 22:43
  • $\begingroup$ We could use coefficients in $\Bbb Z_2$, but I don't have much experience with that personally. So I prefer integer coefficients because of that. $\endgroup$ – Arthur Dec 13 '17 at 22:47
  • $\begingroup$ Okay, I'm just trying to understand then why the assigments of signs we're doing is the correct one so that they cancel out. Still not convinced :P $\endgroup$ – Andy Dec 13 '17 at 22:58
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Singular homology arose by searching very, very, very, very hard for a better proof of the topological invariance of simplicial homology. It had already been known that if $X$, $Y$ are homeomorphic simplicial complexes then their simplicial homology groups are isomorphic; the proof was by something called the "simplicial approximation theorem". A better proof was needed, which would not be so wound up in the details of simplicial complexes, and even better would yield a topological invariant that, when applied to a simplicial complex, is isomorphic to the simplicial homology. The search was successful, and the resulting topological invariant was singular homology.

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