# Does homotopy invariance of homology follow from the structure of the simplex category $\Delta$?

Explicitly: Let $\Delta$ denote the simplex category, and $\mathscr{C}$ any small category, and fix a functor $F:\Delta \rightarrow \mathscr{C}$ such that $F\Delta^0$ is terminal. Also, assume $\mathscr{C}$ has products.

Using $F$, we can define a homology theory in $\mathscr{C}$ letting $C^n(X) = \mathbb{Z}(Hom(F\Delta^n,X))$, the free abelian group generated by morphisms from $F\Delta^n$ to $X$. The boundary map is constructed from the face maps in $\Delta$, and one can prove that $\partial^2 = 0$- this is exactly what is done to find singular homology of topological spaces.

Now, suppose $f,g:X \rightarrow Y$ are maps in $\mathscr{C}$. We say they are homotopy equivalent if there is a map $h:F\Delta^1 \times X \rightarrow Y$ such that $h_0 = f$ and $h_1 = g$. Here $h_0 = X \rightarrow F\Delta^0 \times X \stackrel{F\delta^0 \times 1_X}{\longrightarrow} F\Delta^1 \times X \stackrel{h}{\rightarrow} Y$, where $\delta^0:\Delta^0 \rightarrow \Delta^1$ is a face map. $h_1$ is defined similarly.

My question is: If $f$ and $g$ are homotopic, does it follow that the induced maps of chain complexes $C^{\bullet}(X) \rightarrow C^{\bullet}(Y)$ are homotopic as well?

• The answer to your question is yes. – Daniel Barter Feb 14 '15 at 1:03
• Take the case of singular homology in the category of "nice" topological spaces, and take the corresponding proof (say Lemma II.8.3 from MacLane's Homology) that a homotopy of functions induces a chain-homotopy. What is the non-obvious part of that proof that might not immediately generalize? I think acyclic models is of use here. – Chris Gerig Feb 14 '15 at 1:33
• Chris.. looking at Hatcher, I think you're right. The key is the decomposition of $F\Delta^n \times F\Delta^1$ into $\Delta^{n+1}$s, ie, the collection of maps $\phi_i:F\Delta^{n+1} \rightarrow F\Delta^{n} \times F\Delta^1$. But these maps are just images under $F$ of maps in $\Delta$, and so the whole proof should be unchanged. – Alex Zorn Feb 14 '15 at 2:01
• @Chris: there are no niceness conditions necessary. – Qiaochu Yuan Feb 14 '15 at 8:56
• Alex and Qiaochu, this bring me back to our 1st year officemate days, good times! – Chris Gerig Feb 14 '15 at 20:44

Chris's comment suggests that very little about the target category $C$ is being used in the standard argument, but I still think there's something interesting to check, namely what exactly is being used. The question concerns first of all a "singular chains" functor

$$F_{\bullet}: X \mapsto \left( \Delta^n \mapsto \mathbb{Z}[\text{Hom}(F \Delta^n, X)] \right)$$

from $C$ to the category $\widehat{\Delta}$ of presheaves of abelian groups on $\Delta$ (equivalently, of simplicial abelian groups). After taking singular chains, the rest of the argument will take place almost entirely in $\widehat{\Delta}$.

The question concerns second of all a notion of homotopy in $C$ coming from taking $F \Delta^1$ to be the interval object. The only relevant data in such a homotopy is the induced map

$$F_{\bullet}(F \Delta^1) \times F_{\bullet}(X) \to F_{\bullet}(Y)$$

in $\widehat{\Delta}$. This is the only place where we use that $C$ has finite products and that $F_{\bullet}$, by construction, preserves them. This is also the only place where we use that $F_{\bullet}(X)$ and $F_{\bullet}(Y)$ have anything to do with $C$. Next, note that for every $n$ we have natural maps

$$\Delta^n \to F_{\bullet}(F \Delta^n)$$

(where $\Delta^n$ denotes the corresponding representable presheaf of abelian groups on $\mathbb{Z}[\Delta]$) which arise as follows. By definition, a map $\Delta^n \to F_{\bullet}(F \Delta^n)$ is an element of $\mathbb{Z}[\text{Hom}(F\Delta^n, F\Delta^n)]$. But there is a distinguished such element, namely $\text{id}_{F \Delta^n}$. These maps in fact organize themselves into a natural transformation from the Yoneda embedding $\Delta \to \widehat{\Delta}$ to $F_{\bullet}(F(-))$, which is (a restriction of) the unit map of an adjunction. This implies in particular that they respect the face maps $\Delta^0 \to \Delta^1$.

Now it's clear that pulling back along these natural maps gives a natural transformation from homotopies involving $F_{\bullet}(F \Delta^1)$ to simplicial homotopies

$$\Delta^1 \times F_{\bullet}(X) \to F_{\bullet}(Y).$$

Checking that this natural transformation respects sources and targets as defined in the OP is the only place where we use that $F \Delta^0$ is the terminal object.

The remaining question is why simplicial homotopies of simplicial abelian groups induce chain homotopies on the corresponding chain complexes. This is where the work in Hatcher's proof is, and it has absolutely nothing to do with $C$ or $F$. It exhibits part of the monoidal Dold-Kan correspondence, namely the part having to do with the Eilenberg-Zilber map.

• You write $\times$ but work in the category of simplicial abelian groups. Do you mean $\otimes$ instead? Either way, what does $\Delta^1 \times (-)$ mean? – Zhen Lin Feb 14 '15 at 10:18
• @Zhen: I mean the pointwise product (where $\Delta^1$ is regarded as a representable presheaf), but "map" means bilinear map, so yes, feel free to assume I mean $\otimes$ instead. – Qiaochu Yuan Feb 14 '15 at 10:19
• Incidentally, I think a much stronger statement should be true. $F_{\bullet}$ factors through a "singular simplicial set" functor $S_{\bullet}$, and I think golem.ph.utexas.edu/category/2007/05/… implies that $S_{\bullet}(F \Delta^{\bullet})$ is a resolution of $\Delta^{\bullet}$ as a cosimplicial simplicial set (oof). – Qiaochu Yuan Feb 14 '15 at 10:20
• Yes, I'd rather think about the "singular simplicial set" than pass directly to chains. Then it's much easier to see that simplicial homotopies go to simplicial homotopies, and everything else is as for simplicial homology. I don't follow what you say about cosimplicial resolutions, though. – Zhen Lin Feb 14 '15 at 10:23
• I mean we can use the objects $S_{\bullet}(F \Delta^n)$ not only to talk about an $F$-based notion of homotopy but an $F$-based notion of higher homotopy, and I think this is the same as the notion of higher homotopy you get from the $\Delta^n$. In particular, $S_{\bullet}(F \Delta^n)$ should be contractible for all $n$, which I don't think follows from what I've said but should follow from the fact that $S_{\bullet}(F(-))$ is (the restriction of) a monad, if the blog post above says what I think it says. – Qiaochu Yuan Feb 14 '15 at 10:26