# Simplicial homology for n-simplex

I've just started to study homology theory. And I'm trying to calculate all $H_n(\Delta_N)$ for some $N$. I know that the number of $m$-simplex in $N$-simlex is $b_{N,m}={N+1 \choose m+1}=\frac{(N+1)!}{(m+1)!(N-m)!}$.

So $C_0=\mathbb Z^{N+1}$, $C_1=\mathbb Z^{\frac{N(N+1)}{2}}$, ... , $C_N=\mathbb{Z}$, $C_{N+1}=0$, ...

And here I have some misunderstanding. As I know I shoud calculate $H_n(\Delta_N)=Ker(\partial_n)/Im(\partial_{n+1}))$. In opposite to standard examples I don't know what I should do next. Maybe there is some another ways to solve this problem?

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It's less important to count the number of $m$-simplices and more important to know where the boundary maps $C_{m+1} \to C_m$ send them! Perhaps if you know what the answer is it will be easier to see what you need to do: so as a hint, I'll tell you that $H_0 \cong \mathbb{Z}$ and $H_n = 0$ for $n > 0$. – Zhen Lin Sep 17 '13 at 16:49
@Zhen Lin Thanks for your comment! I understand why $Ker(\partial_0)=\mathbb{Z^{n+1}}$, but I don't get why $Im(\partial_1)=\mathbb{Z^{n}}$. – Oiale Sep 17 '13 at 18:34
Use the fact that homology groups are homotopy invariant. Then compute homology of a point. – studiosus Sep 17 '13 at 20:25

It's allmost trivial if you know the following fact of homology, wich you can found in any standard text (Hatcher for example):

Theorem. Homology groups are homotopy invariant, that is, if $X$ is homotopic to $Y$ then $H_n(X)\cong H_n(Y)$

So now you only need to see that every convex set is contractible.

Lemma. Let $X$ be a convex set, then $X$ is contractible.

Proof. Let $x_0$ be a point and define a map $F:X\times I \to X$ by $F(x,t) = tx_0 + (1-t)x$ this map is obviously continuous and is well defined because $X$ is convex. You can easily see that this give us the homotopy between the identity and the constant function so $X$ is contractible.

By the previous theorem and the fact that the homology of a point is $0$ for every $n>0$ and $\mathbb{Z}$ in dimension $0$ you get your result.

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