Homotopy of singular $n$-simplices I was wondering if there's any way to fit in homotopy into the definition of singular homology. Assume that two singular $n$-simplices are homotopic as maps, does this relate them in any way as elements of the homology group $H_n$?
I guess we should at least get some information for $H_1$ as homotopic loops would generate the same thing in the fundamental group, so when taking the abelianization, they would definitely represent the same element in $H_1$ as singular $1$-simplices.
How much can we say in general? Are there any interesting cases where this provides us with interesting information?
 A: $\Delta^n$ doesn't really carry any interesting homological information, since simplices are contractible. 
If we add the restriction that the boundary of the simplex maps to a single point, the analog of a loop (indeed, a singular cycle), we in fact see that $\Delta^n / \partial \Delta^n \cong S^n$, so maybe we should be looking at maps out of spheres. Alternately, you could just be thinking about $\partial \Delta^n = S^{n-1}$ to begin with.
In fact, maps from spheres are much more fruitful. A map $f : S^n \to X$ gives an induced map $f_* : H_n(S^n) \to H_n(X)$. Homotopic maps $f,g : S^n \to X$ give equal maps $f_*,g_*$. Notice that $H_n(S^n) = \mathbb{Z}$, and so $f_*$ picks out an element of $H_n(X)$. So in this sense, homotopic maps of spheres pick out equal elements of homology.
If you know about higher homotopy groups, this defines a natural homomorphism $\pi_n(X) \to H_n(X)$ called the Hurewicz map.
This map has a very useful property: if $\pi_1$ is nonvanishing, then the map $\pi_1(X) \to H_1(X)$ is the abelianization homomorphism. If $\pi_k$ vanishes for $k < n$, and $n \ge 2$, then $\pi_n(X) \to H_n(X)$ is in fact an isomorphism! This is a very useful tool for computing the first nonvanishing homotopy group of a space, since homology is typically much easier to compute than the homotopy groups are.
