TITLE: Simplicial homology for n-simplex
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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.