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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?

Eric Auld
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1 Answers1

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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 $\sum a_i \sigma_i'$.

Eric Auld
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Eric Wofsey
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  • $K$ is then a simplicial complex without boundary? – Eric Auld Oct 17 '15 at 20:06
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    It's a $\Delta$-complex; it need not be a simplicial complex (at least with its obvious simplicial structure)--for instance, in your case of $S^1$, you are giving it the simplicial structure with just one edge and one vertex, which is certainly not a simplicial complex. I don't know what it means for a simplicial complex to have (or not have) a boundary. – Eric Wofsey Oct 17 '15 at 20:08
  • Excuse me, I meant a $\Delta$-complex. Does it make sense to say that a $\Delta$-complex has no boundary? Perhaps the definition would be there is an upper bound on the dimension of simplices and the sum of their boundaries is zero? – Eric Auld Oct 17 '15 at 20:15
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    I guess you could say that a finite $\Delta$-complex has no boundary if you can assign "orientations" to each of its top-dimensional simplices such that the sum of their boundaries (with signs provided by the orientations) vanishes. That need not be true of my $K$, because you only know that the sum $\sum a_ix_i$ with the given coefficients $a_i$ has trivial boundary, and the $a_i$ might not be $\pm1$. If you want to, however, you could instead construct $K$ with $|a_i|$ different copies of the simplex $x_i$, so that you could assume all the coefficients $a_i$ are $\pm 1$. – Eric Wofsey Oct 17 '15 at 20:28
  • Sorry for the vague question, but does this give us any particular corresponding insight into cohomology? I don't see what "homotopic cochains" would mean. (Plainly your answer shows that cocycles take equal values on chains which are homotopic through the types of maps we discussed.) – Eric Auld Feb 07 '16 at 22:19
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    Well, I don't know of any notion of "homotopic cochains" that is so nicely geometrical. But you can still talk about cohomology classes which are pulled back from the same class on a space $K$ by two homotopic maps $X\to K$. – Eric Wofsey Feb 08 '16 at 04:23
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    There is no longer a canonical space $K$ that is geometrically associated to any particular cocycle, as there is for homology. But in some sense there is something even nicer: there is a single space $K$ with a single cohomology class such that every class in $H^n$ can be obtained by pulling back via a (unique up to homotopy!) map $X\to K$, namely $K=K(A,n)$ where $A$ is the coefficient group. – Eric Wofsey Feb 08 '16 at 04:24
  • Thanks! Regarding homology, I think the analogous statement for relative $(X,A)$ cycles is "if two relative cycles are homotopic via homotopies which a) have the property that we discussed above, and b) keep any boundaries which lie in $A$ inside $A$, then they are homologous." To prove this, I think I would repeat the construction of $K$ as above, and look at maps $(K,L) \to (X,A)$, where $L$ is the image of the boundary of $c$. (Since $c$ may now not be a cycle, only a relative cycle.) Would you modify that? – Eric Auld Feb 09 '16 at 00:16
  • Sounds good to me. In fact, you don't even need all the boundary faces that are in $A$ to stay in $A$, only the ones that fail to cancel out when you take the boundary of the entire chain. – Eric Wofsey Feb 09 '16 at 00:20