# cohomology group of $SO(n)$

I am computing the Alexander-Spanier cohomology $H^i(SO(n),\mathbb{Z})$. I embedded $SO(n)$ into $R^{n^2}$. Since the embedding $i$ is a monomorphism, the induced group homomorphism $i^*$ is an epimorphism. Since $R^{n^2}$ is homotopic to a point, $H^i(R^{n^2})=0 , \forall i \in \mathbb{Z}^+$. That gives us $H^i(SO(n))=0, \forall i \in \mathbb{Z}^+$.

I don't know where I am wrong, can anyone give any suggestions?

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Since $SO(n)$ is a manifold, its Alexander-Spanier cohomology is just its singular cohomology. This can be computed inductively via the Serre spectral sequence for the fibration $SO(n) \rightarrow SO(n+1) \rightarrow S^n$. ($SO(n+1)$ acts transitively on $S^n$, and the stabilizer of any point is the copy of $SO(n)$ acting on its equatorial $S^{n-1}$.) – Aaron Mazel-Gee May 14 '12 at 1:29
By the way, homotopy-theoretic invariants like cohomology groups are just that: quantities that are invariant under homotopies. So in general, they're not going to know about properties like "monomorphic". – Aaron Mazel-Gee May 14 '12 at 1:34

It is not true that if $i:M\to N$ is an embeding you can deduce that dually $i^\ast:H^r(N,\mathbb Z)\to H^r(M,\mathbb Z)\$ is an epimorphism.
For example, the embedding of the circle in the plane $i:S^1\to \mathbb R^2$ gives rise to $i^\ast: H^1(\mathbb R^2,\mathbb Z)=0\to H^1(S^1,\mathbb Z)=\mathbb Z$, which clearly cannot be an epimorphism.