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I'm reading through this proof by Hatcher and I don't understand how this is true. enter image description here

How can you divide $C$ and $I_j$ to guarantee you are mapping into a single $U_{\alpha}$. As it makes sense formally, but surely the proof you would need to use some continuity or have to do a special construction. The only thing I can think of is arguing that you can make a distinct covering of the space

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The result used in this type of argument is called Lebesgue's number lemma: – Justin Young May 6 '12 at 22:11
up vote 2 down vote accepted

Since nobody actually answered, I'll expand on Justin's comment.

If $\{U_\alpha\}$ is a cover of $B$, then $V_\alpha:=G^{-1}(U_\alpha)$ is a cover for $I^n\times I$. Since $I^n\times I$ is a compact metric space, the Lebesgue number lemma applies, giving us a number $\delta$ such that every set of diameter less than $\delta$ lies in one of the $V_\alpha$'s. In particular, this means we can divide $I^n$ into cubes $C_k$ of diameter less than $\delta/2$ and $I$ into intervals $I_j$ of length less than $\delta/2$ so that $C_k\times I_j$ has diameter less than $\delta$ and thus lies entirely in some $V_{\alpha_{k,j}}$. But this means that $G(C_k\times I_j)\subseteq U_{\alpha_{k,j}}$ which is what we wanted.

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