I'm trying to solve the following problem:
Show that for a degree 1 map $f: M \rightarrow N$ of connected, closed, orientable manifolds, the induced map $f_*: \pi_1(M) \rightarrow \pi_1(N)$ is surjective and hence also the map $f_*: H_1(M) \rightarrow H_1(N)$ is surjective.
I didn't struggle with proving the first map is surjective, here's my proof:
For each subgroup of $\pi_1(N)$ we have a covering space of $N$ corresponding to it. Let $\widetilde{N}$ be the covering space corresponding to $im(f_*) \subset \pi_1(N)$.
Then we have a lift $\widetilde{f}: M \rightarrow \widetilde{N}$ and a diagram that shows $p \circ \widetilde{f} = f$. So now we have $1 = deg(f) = deg(p \circ \widetilde{f}) = deg(p)\cdot deg(\widetilde{f})$.
(Where $p$ is the projection from $\widetilde{N}$ onto $N$.)
Then since $deg(p)$ and $deg(\widetilde{f})$ are integers, we have they both are equal to $-1$ or $1$. Let's say that they're equal to $1$. But this implies that $p$ is a $1$-sheeted cover. Therefore, $p$ is a homeomorphism.
But this means $\pi_1(\widetilde{N}) = \pi_1(N)$, so $im(f_*) = \pi_1(N)$ so $f_*$ must be a surjection.
I'm not sure how the abelianizations of the fundamental groups extend this to a surjection from $H_1(M) \rightarrow H_1(N)$.
Can anyone help me with this part?
Best Answer
This is basically just algebra.
You see that $\pi_1(M) \to \pi_1(N) \to H_1(N)$ is surjective as composition of surjective maps. Now $H_1(N)$ is abelian, hence this composition factors through the abelianization of $\pi_1(M)$ which is $H_1(M)$ and this factorization yields the map $H_1(M) \to H_1(N)$ you are asking for. So surjectiveness follows on the nose.
It might be easier to see the result, if you write the first homologies down as quotients of the fundamental groups.