# Finite simplicial complex can be viewed as a subcomplex of a simplex

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?

• It has finitely many vertices. – Lord Shark the Unknown Dec 4 '19 at 17:22
• @Lord Shark the Unknown Can you give a little bit more explanation? – Quadr Dec 4 '19 at 17:24

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 $$K$$.