I have a question about the homology of the real projective space $\mathbb{R} P^2$ with which I'm having some trouble:
Let $f: \mathbb{R}P^2 \rightarrow \mathbb{R}P^2$ be a map which induces an isomorphism in homology. Why is $f$ surjective?
I know that the homology of $\mathbb{R}P^2$ is as follows:
$H_0(\mathbb{R}P^2) = \mathbb{Z}$,
$H_1(\mathbb{R}P^2) = \mathbb{Z}/2\mathbb{Z}$
$H_i(\mathbb{R}P^2) = 0$ for any $i \neq 0,1$.
This is true for any homology theory (by the use of cellular homology).
What I've tried is to lift $f$ to a map $g: S^2 \rightarrow \mathbb{R}P^2$ and then arrive at a contradiction to $f$ being an isomorphism in homology when assuming that the degree of $g$ is zero. From this it would follow that $deg(g) \neq 0$, hence $g$ is surjective (and therefore also $f$). However this argument doesn't work because $H_2(\mathbb{R}P^2) = 0$.
Any help would be greatly appreciated.
EDIT: Even though my initial search was fruitless I just now found a more or less satisfying answer: Real projective plane: $f_*$ isomorphism $\implies f$ surjective
$\Bbb RP^2$ is a Moebius strip with a disk attached along its boundary. You can convince yourself of this by considering $\Bbb RP^2$ as the quotient space of a hemisphere $$S^2_+=S^2\cap\lbrace z\geq 0\rbrace$$ (with $S^2=\lbrace(x,y,z)\in\Bbb R^3\mid x^2+y^2+z^2=1\rbrace$) under the equivalence relation identifying a point of the equator (i.e. with $z=0$) to its opposite. In any case, $\Bbb RP^2\setminus\lbrace\mathrm{pt}\rbrace$ is homeomorphic to a Moebius strip, and thus homotopy equivalent to a circle. Since homology groups are homotopy invariants, and $H_1(\Bbb RP^2\setminus\lbrace\mathrm{pt}\rbrace)\simeq H_1(S^1)$ is free abelian on one generator, it isn't isomorphic to $H_1(\Bbb RP^2)$.