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Simplicial Approximation Theorem for maps roughly states:

If $X$ and $Y$ are two finite simplicial complexes and $f:|X|→|Y|$ is a continuous map between their geometric realizations, then there exists a subdivision $X'$ of $X$ and a simplitial map $g: X' \to Y$ such that $|g|$ is homotopic to $f$.

Can someone help me how to show (by using this theorem) that the set $[X,Y]$ of homotopy classes $f: X \to Y$ is at most countably infinite.

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The point here is that one can choose $X'$ to be an iterated barycentric subdivision $X^{(k)}$ of $X$. There are finitely many simplicial maps from $X^{(k)}$ to $Y$ for a given $k$, so countably many overall.

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  • $\begingroup$ Thanks, but what do you mean by "iterated" subdivision? Using App. Theorem I only know that there is "some" subdivision satisfying certain properties.. $\endgroup$ – edward_scissorhands Apr 24 '17 at 17:07
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    $\begingroup$ There is a procedure called simplicial subdivision. If one does this $k$ times to $X$ we get $X^{(k)}$. The proof of simplicial approximation shows that one can take $X'$ to be any sufficiently fine subdivision of $X$. $\endgroup$ – Angina Seng Apr 24 '17 at 17:09

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