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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?).

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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
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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.

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