# Surjective map on real projective space

If $f$ is map of $\mathbf{RP}^2$, inducing an isomorphism in homology, then $f$ is surjective. How can one prove this, maybe using some kind of CW complex structure on it or how?

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By a map of $\Bbb R \mathrm P^2$, do you mean $\Bbb R \mathrm P^2 \to \Bbb R \mathrm P^2$? – Ayman Hourieh Feb 2 '14 at 23:45
Exact duplicate of math.stackexchange.com/questions/655791/… – Olivier Bégassat Feb 3 '14 at 0:23

Suppose $f$ misses a point $p \in \Bbb R \mathrm{P}^2$. Then $f$ factors through $Y = \Bbb R \mathrm{P}^2 - \{p\}$: $$\Bbb R \mathrm{P}^2 \xrightarrow{g} Y \xrightarrow{i} \Bbb R \mathrm{P}^2$$
$Y$ is homotopy equivalent to $S^1$. Consider the fundamental polygon of $\Bbb R \mathrm{P}^2$ to see this. This forces $g_* : H_1(\Bbb R \mathrm{P}^2) \to H_1(Y)$ to be trivial since $H_1(\Bbb R \mathrm{P}^2) \cong \Bbb Z_2$ and $H_1(Y) \cong \Bbb Z$.
It follows that $f_* = i_* \circ g_*$ cannot be an isomorphism.
Did you considered the case $f$ missing more than $1$ point? It seems the proof is not complete. – Bombyx mori Feb 3 '14 at 1:01
@user32240 It doesn't matter whether $f$ misses more points or not as long as it misses at least one. The proof doesn't assume that $g$ is surjective. – Ayman Hourieh Feb 3 '14 at 1:02
@amoreacceptablename: I think the proof would be complete if Ayman mentioned $g$ is continuous. – Bombyx mori Feb 4 '14 at 0:13