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I know that, for every $\Delta$-complex $X$, there is a canonical isomorphism $\phi_n : H_n ^\Delta (X) \to H_n (X) $, where $H^\Delta _n (X)$ is the $n$-th simplicial homology group, and $H_n (X)$ is the $n$-th singular homology group.

For simplicial $\Delta$-complexes, we have also the notion of order-preserving simplicial maps, and every simplicial map $f:X\to Y$ induces a morphism $f^\Delta _{\bullet_*}$ in simplicial homology $H^\Delta _\bullet $.

My question is: are the morphisms $f^\Delta _{\bullet_*}$ and $f _{\bullet_*}$ (the induced morphism in singular homology) "the same" via the canonical isomorphisms? I believe that this is true, but I'm having troubles proving this, because if $\sigma$ is a $n$-simplex of $X$ and $f$ sends $\sigma$ in a simplex of $Y$ such that $\dim f(\sigma) < \dim \sigma$, then $ f_n (\sigma)=0 $, but if we denote $c_\sigma : \Delta ^n \to X $ the characteristic map of $\sigma$ in the $\Delta$-complex structure of $X$, then $ f\circ c_\sigma :\Delta^n \to Y$ is a well defined $n$-singular simplex of $Y$.

Thank you.

Eric Wofsey
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Ervin
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1 Answers1

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Yes, this is true. Here's one way to prove it. For $0\leq j<n$, write $s_j^{n-1}:\Delta^n\to\Delta^{n-1}$ for the map induced by the order-preserving surjection $\{0,\dots,n\}\to\{0,\dots,n-1\}$ that maps both $j$ and $j+1$ to $j$. Say that a singular simplex $\Delta^n\to X$ is degenerate if it factors through $s_j^{n-1}$ for some $j$. Note that the boundary of a degenerate simplex is a linear combination of degenerate simplices: all but possibly two of its faces are degenerate (the two exceptions being the faces corresponding to omitting the vertices $j$ and $j+1$), and those two faces cancel out. So if we write $D_n(X)\subseteq C_n(X)$ for the span of the degenerate simplices, $D_\bullet(X)$ is a subcomplex of $C_\bullet(X)$. Write $N_\bullet(X)=C_\bullet(X)/D_\bullet(X)$.

Now note that given an order-preserving simplicial map $f:X\to Y$, the induced maps $C^\Delta_\bullet(X)\to C^\Delta_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)$ and $C^\Delta_\bullet(X)\to C_\bullet(X)\to C_\bullet(Y)\to N_\bullet(Y)$ are equal, since the $n$-simplices $\sigma$ that you're worried about have the property that $f\circ c_\sigma$ is degenerate. So to show that $f^\Delta_{\bullet_*}=f_{\bullet_*}$, it suffices to show that the map $C_\bullet(Y)\to N_\bullet(Y)$ induces isomorphisms on homology. By the long exact sequence in homology associated to the short exact sequence $0\to D_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)\to 0$, it suffices to show that $D_\bullet(Y)$ has trivial homology.

We can show this by constructing a chain homotopy. Given a degenerate $n$-simplex $\sigma:\Delta^n\to Y$, let $j(\sigma)\in\{0,\dots,n-1\}$ be the least $j$ such that $\sigma$ factors through $s_j^{n-1}$. Now define $H:D_n(Y)\to D_{n+1}(Y)$ by $H(\sigma)=(-1)^{j(\sigma)}\sigma\circ s^n_{j(\sigma)}$. An elementary computation then shows that $H\partial+\partial H:D_n(Y)\to D_n(Y)$ is the identity map. It follows that $D_\bullet(Y)$ has no homology.

To give a sense of the computation $H\partial+\partial H=1$, let's show what happens when you take a $3$-simplex $\sigma$ with $j(\sigma)=1$; the general case works very similarly. Let's write $\sigma=[a,b,b,c]$; all the simplices built from $\sigma$ will be written using similar expressions with the obvious meaning (here $a$, $b$, and $c$ are the vertices of $\sigma$, with $b$ repeated since it factors through $s^2_1$). We then have $H(\sigma)=-[a,b,b,b,c]$, so $$\begin{align}\partial H(\sigma)=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,c]+[a,b,b,c]-[a,b,b,b]\\=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,b].\end{align}$$ On the other hand, $\partial\sigma=[b,b,c]-[a,b,c]+[a,b,c]-[a,b,b]=[b,b,c]-[a,b,b]$. We have $H([b,b,c])=[b,b,b,c]$ and $H([a,b,b])=-[a,b,b,b]$. Thus we get $$H(\partial\sigma)=[b,b,b,c]+[a,b,b,b].$$ When we add together $\partial H(\sigma)$ and $H(\partial\sigma)$, all the terms cancel except $[a,b,b,c]$, which is just $\sigma$.

Eric Wofsey
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  • This is reminiscent of the normalized bar resolution having the same homology as the unnormalized bar resolution. – Pedro Dec 26 '15 at 02:18
  • @PedroTamaroff: Yes, both are special cases of the fact that for any simplicial object in an abelian category, the associated chain complex is homotopy-equivalent to the "normalized" chain complex you get by modding out the subcomplex of degenerates. – Eric Wofsey Dec 26 '15 at 02:24
  • Aha, that doesn't come as a surprise. :) – Pedro Dec 26 '15 at 02:27