Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $X$ is a CW-complex and $f:X\rightarrow X$ a cellular map. Then why it induces a map of chain complexes $f_*:C_*(X)\rightarrow C_*(X)$. (Why it commutes with the differential?).

share|improve this question
Have you looked in any algebraic topology textbook that offers a proof of this? Bredon, Hatcher, May, Whitehead, etc.. Once you interpret what the chain complex is, your question boils down to the fact that the degree of a map $S^n \to S^n$ is multiplicative under composition. The proof of this is fairly simple. There are transversality arguments, and homological arguments. –  Ryan Budney Oct 3 '11 at 22:29

1 Answer 1

Let me give the sketch of an homological argument.

Let $X$ be a non-empty finite CW-complex of dimension $k$, and let $$\emptyset=X^{-1}\subsetneq X^0\subseteq X^1\subseteq\cdots\subseteq X^k=X$$ be the increasing sequence of skeletons of $X$. For each space $Y$, let $\bar S_\bullet(Y)$ be the reduced singular complex of $Y$. For each $p$ let $$F^p\bar S_\bullet(X)=\bar S_\bullet(X^p).$$ This defines an increasing filtration on the complex $\bar S_\bullet(X)$, and we can consider the corresponding spectral sequence $E=E(X)$. The $0$th page of $E$ has $$E^0_{p,q}=\frac{F^p\bar S_{p+q}(X)}{F^{p-1}\bar S_{p+q}(X)}=\frac{\bar S_{p+q}(X^p)}{\bar S_{p+q}(X^{p-1})},$$ and this is the the degree $p+q$ part of the reduced relative complex $\bar S_{p+q}(X^p,X^{p-1})$. The differential on $E^0$ is induced by that of $\bar S_\bullet(X)$. It follows at once from this that $E^1_{p,q}=\bar H_{p+q}(X^p,X^{p-1})$.

Now, since $X^p$ is obtained by attaching $p$-cells to $X^{p-1}$, a standard computation shows that $E^1_{p,q}=0$ if $q\neq0$. This implies that the spectral sequence degenerates at $E^2$, and —since it converges—, that $\bar H_\bullet(X)$ is the homology of the complex $$\cdots \to \bar H_p(X^p,X^{p-1}) \xrightarrow{\quad d^1_{p,0}\quad } \bar H_{p-1}(X^{p-1},X^{p-2}) \to \cdots$$ In particular, this is a complex. Now a little consideration of commutative diagrams shows that this map $d^1_{p,0}$ is the cellular differential.

Now, suppose that $f:X\to Y$ is a map of two CW-complexes as above which maps each skeleton to the corresponding skeleton. Then the induced map $f_\bullet:\bar S_\bullet(X)\to\bar S_\bullet(Y)$ respects the filtrations on $\bar S_\bullet(X)$ and on $\bar S_\bullet(Y)$, it it in fact induces a morphism $f_{\bullet,\bullet}:E(X)\to E(Y)$ between the corresponding spectral sequences. This statement in particular includes the fact that $f_{\bullet,\bullet}$ commutes with the cellular differential.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.