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I'm trying to do exercise 4 on page 38 in Hatcher. Can you tell me if this is right?

claim: $X$ locally star-shaped, $\gamma$ a path in $X$ then there is a path consisting of a finite number of line segments that is homotopic to $\gamma$

proof:

By assumption $\gamma(0)$ has a star-shaped neighbourhood, let's call it $U_0$. Consider $\gamma(t_1) \in U_0$. Because $U_0$ is star-shaped there exists a straight line between $\gamma(0)$ and $\gamma(t_1)$. Repeat the same for $\gamma(t_1)$ and so on to get a path consisting of straight lines: $$\gamma(0) \rightarrow \gamma(t_1) \dots \rightarrow \gamma(1)$$

This is homotopic to $\gamma$ because every point of it can be connected to $\gamma$ via a straight line.

Thanks for your help!

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I think you need to be a bit more careful: what if $\gamma([0,t_1]) \not\subset U_0$? –  t.b. Aug 31 '11 at 8:39
    
Could I not choose a smaller $t_1$ in that case? –  Matt N. Aug 31 '11 at 8:41
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Sure, but you should mention that. Also, you should probably spell out what exactly the homotopy is and why it is continuous. The idea is of course right but there are a few more details to be checked. –  t.b. Aug 31 '11 at 8:45
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1 Answer 1

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For each $x \in [0,1]$ let $N_{\gamma(x)}$ be a starshaped nbhd of $\gamma(x)$. Because $\gamma([0,1])$ is compact and the $N_{\gamma(x)}$ cover it, you can thin them to a finite sub-cover. The rest should be easy.

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