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Let $f:T^2 \to S^2$ be the map that collapses the 1-skeleton $S^1 \vee S^1$ Compute $f_∗$,$f^*$.

So start with the usual CW strucutre on the sphere (1 0-cell and 1 2-cell) and the torus (1 0-cell, 2 1-cells and 1 2-cell). Now all the boundary maps are zero in the chain complex of the torus. So let us map the 1-skeleton onto the zero cell and map the 2 cell of the torus onto the 2 cell of $S^n$ by the identity map. Since the boundary maps on the torus are 0, we have a continuous $f:T^2 \to S^2$ (I think?).

$f$ is the identity (edit: on the 2-cell), so does this imply that $f^*:H^2(S^2) \to H^2(T^2)$ and $f_*:H_2(T^2) \to H_2(S^2)$ are isomorphisms?

So I guess my answer to the question would be - the maps $f_n$ and $f^n$ are isomorphism's for $n=2$ and zero else?

Edit: Acutally I still have the 0 dimension to worry about as well I guess?

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$f$ is not the identity. –  Rasmus May 13 '11 at 8:32
    
@Rasmus - sorry I meant we map the $n$ cell to the $n$ cell by the identity (so send the 1 skeleton to the 0 cell, and send the 2 cell of the torus onto the 2 cell of the sphere by the identity) –  Juan S May 13 '11 at 8:38
    
Yes, if you use cellular cohomology this should be clear. But note that the map in cohomology goes the other way! (Homology is a covariant functor, cohomology is a contravariant functor.) There's not much to worry about in dimension 0, but it's still worth thinking through if you're confused. –  Aaron Mazel-Gee May 13 '11 at 15:17
    
@Aaron: Oh yes indeed, that is an embarrassing mistake! –  Juan S May 14 '11 at 5:28
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