Just learning about simplicial homology.
Suppose $X$ is a topological space, and suppose I have a $1$ simplex $\sigma_1 : \Delta^1 \to X$ which is itself a cycle, i.e., it descends to a map from $S^1$ into $X$. If I have a homotopy of such maps $F: S^1 \times I \to X$, not necessarily relative to any basepoint, then I can view the homotopy as a map $I \times I \to X$ with two sides identical. If I give this a simplicial structure, I have a two-simplex with boundary $\sigma_2 – \sigma_1$ and I can conclude that in some restricted sense homotopic cycles consisting of a single simplex are homologous.
(Aside: It makes sense that in no longer considering basepoint I have abelianized, I think, because for instance if I look at $S^1 \vee S^1$ with $\pi_1(S^1 \vee S^1) \simeq \langle a \rangle \ast \langle b \rangle$, then I can get from $ab$ to $ba$ by sort of sliding my string through (i.e. homotoping not rel the basepoint).)
It seems like I should be able to do this in greater generality and in higher dimensions (of course not for just any two homotopic simplices), but I'm not seeing how. (One obstacle is that it's not clear to me what a homotopy of chains with more than one term would mean, and in even dimensions, a single simplex can't be a cycle.) It it perhaps a special case in a clever way of the theorem that homotopic maps induce the same maps on homology? Is there a general way, given certain hypothesis, to go from two simplices/chains which are homotopic to those which are homologous?
Best Answer
Yes, this is just a special case of homotopic maps inducing the same map on homology. In particular, there is a 1-simplex $\sigma:\Delta^1\to S^1$ which is itself a cycle, and your cycle $\sigma_1$ is just the image of $\sigma$ under the corresponding map $S^1\to X$. Your argument that $\sigma_1$ and $\sigma_2$ are homologous by triangulating $I\times I$ is literally exactly the usual proof that homotopic maps induce the same map on homology, applied to this special case.
More generally, if $c=\sum a_i \sigma_i$ is a $n$-cycle in a space $X$, you can form a space $K$ which is obtained from $n$-simplices $x_i$ by gluing the $j$th face of $x_i$ to the $j'$th face of $x_{i'}$ whenever the $j$th face of $\sigma_i$ is equal to the $j'$th face of $\sigma_{i'}$ as maps $\Delta^{n-1}\to X$. Then $x=\sum a_i x_i$ is an $n$-cycle in $K$, and there is a map $f:K\to X$ which sends $x_i$ to $\sigma_i$ and so sends $x$ to $c$. If $f$ is homotopic to a map $g:K\to X$, then $c$ is homologous to the cycle $g_*(x)$. Concretely, a homotopy from $f$ to some other map $g$ consists of a homotopy from each simplex $\sigma_i$ to a new simplex $\sigma_i'$ such that each stage of the homotopy preserves any coincidences among the boundary faces of the $\sigma_i$ (in your example, this corresponds to the fact that $\sigma_1$ and $\sigma_2$ need to be homotopic through 1-simplices that keep the two vertices identified). Whenever you have such homotopies, you can conclude that $\sum a_i\sigma_i$ is homologous to $\sum a_i \sigma_i'$.