# Explain why this homotopy works?

It's page 351, but I'm having trouble with the line highlighted.

The highlighted part makes no sense to me. I can see why $f_0=f$ as you would do this $f_0=(1-0)f+0g=f$, but then how do you get $f_1|K_1=g|K_1$. I was thinking that $f_1|K_1=(1-\varphi)f+\varphi g$?

So how do you get that to be equal to g. I know g and f agree on vertices. But, I don't see how you get it. The before this bit is fine, but then this bit confuses me.

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Since $\phi$ is linear on simplices and constant on vertices in $K_1$, we must have $\phi$ constant (and equal to 1) on all of $K_1$, which is what you want.

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How does the f disappear. If we $\phi$ being constant, then I don't know why the map isn't equal to f+g. So I was worrying about why the f disappears. – simplicity Apr 23 '12 at 18:23
$\phi$ is constantly 1 so $(1 - \phi)f$ is 0 and $\phi g$ is $g$. – user29743 Apr 23 '12 at 18:24

Without answering this question on Hatcher's proof, I mention that the proof of the Cellular Approximation Theorem in Section 7.6 of my book "Topology and groupoids" uses no simplicial subdivisions, but only cubical ones, and you can find the same proof in older editions of this book, with different titles, which may be in your library. I took it long ago from handwritten notes of J.F. Adams. (The proof given restricts to the case of finite cell complexes, but the topology in the general case is arranged so that the same proof goes through!) Here is the key step.

The following statements are true for each $n \geqslant 1$.

$\alpha(n)$ Any map $S^r \to S^n$ with $r < n$ is inessential.

$\beta(n)$ Any map $S^r \to S^n$ with $r < n$ extends over $E^{r+1}$.

$\gamma(n)$ Let $B$ be path-connected and let $Q$ be formed by attaching a finite number of $n$-cells to $B$. Then any map $$(E^r,S^{r-1}) \to (Q,B)$$ with $r<n$ is deformable into $B$.

The proof is by induction by means of the implications $$\gamma(n) \Rightarrow \alpha(n) \Leftrightarrow \beta(n) \Rightarrow \gamma(n+1)$$ the only difficult step being the proof of $\beta(n) \Rightarrow \gamma(n+1)$.

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