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I am reading the proof of the Simplicial Approximation Theorem (2C.1) in Hatcher.

In the first paragraph of the proof, Hatcher says that:

Choose a metric on $K$ that restricts to the standard Euclidean metric on each simplex of $K$. For example, $K$ can be viewed as a subcomplex of a simplex $\Delta ^N$ whose vertices are all the vertices of $K$, and we can restrict a standard metric on $\Delta ^N$ to give a metric on $K$.

Here $K$ is just a finite simplicial complex.

I understand that the second sentence clearly implies the first, but I cannot understand why we can view $K$ as a subcomplex of a simplex $\Delta ^N$. Any help?

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  • $\begingroup$ It has finitely many vertices. $\endgroup$ – Lord Shark the Unknown Dec 4 '19 at 17:22
  • $\begingroup$ @Lord Shark the Unknown Can you give a little bit more explanation? $\endgroup$ – Quadr Dec 4 '19 at 17:24
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The key idea is that in a simplicial complex (unlike in, say, a $\Delta$-complex), each simplex is uniquely determined by its vertices (this is part of the definition of a simplicial complex). So, since $K$ has finitely many vertices, say $N - 1$, consider the simplex $\Delta^n$, and identify the $N - 1$ vertices of $\Delta^N$ with the $N - 1$ vertices of $K$.

Now let's think about how to identify the $k$-simplices of $K$ with $k$-simplices of $\Delta^N$. For any $k + 1$ vertices in $K$, there might or might not be a $k$-simplex in $K$ with those vertices, but there's at most one $k$-simplex in $K$ with those vertices (since $K$ is a simplicial complex). There is also exactly one $k$-simplex in $\Delta^N$ with those $k + 1$ vertices. So, if $K$ has a $k$-simplex with those vertices, include the $k$-simplex in $\Delta^n$ with these vertices; if it doesn't, don't. Repeating this process and taking all simplices in $\Delta^n$ that correspond to simplices in $K$, we obtain a subcomplex of $\Delta^n$ which is homeomorphic to our original complex $K$.

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