Demystifying math: How someone you could have come up with singular homology? 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:

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*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.
 A: 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.
A: 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.
A: 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.
