# Prove the map is null homotopy

Prove that every continuous map $f:P^2\to S^1$ is null homotopy.

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What have you tried so far? Is $P^2$ the projective plane? In that case, $\pi_1(P^2)=\mathbb{Z}/2$. HINT: What are the maps $\mathbb{Z}/2 \to \mathbb{Z}$? – Fredrik Meyer May 9 '12 at 9:38
@FredrikMeyer's suggestion is what you need. See also a related post. – bgins May 9 '12 at 9:40
I'm learning covering space and this is an exercise for that.I think I should use things related to covering space but $\pi_1(\mathbb P^2)=\mathbb Z_2$...ah,maybe I see how to prove it. – Jiangnan Yu May 9 '12 at 9:44
@JiangnanYu No, because all contractible spaces have trivial fundamental group. – Matthew Pressland May 9 '12 at 10:18
If you are having trouble characterizing $\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}_2,\mathbb{Z})$, you should probably go back and brush up on commutative algebra before you continue studying algebraic topology ... – Neal May 9 '12 at 16:16

As Jim pointed out, it's not enough to show that $f_*$ induces the trivial map on fundamental groups. To finish it, we can proceed as follows. Let $p\colon \mathbb R\to S^1$ be the covering projection. Since $f_*(\pi_1(P^2))=0$ (as Fredrik showed), it's clearly contained in $p_*(\pi_1(\mathbb R))=0$. So, this means there is a map $\tilde f\colon P^2\to\mathbb R$ such that $f=p\tilde f$ (Proposition 1.33 in Hatcher). Now, $\mathbb R$ is contractible so we have a homotopy $H\colon P^2\times I\to \mathbb R$ such that $H_0=\tilde f$ and $H_1$ is constant at some $x_0$. But then $pH$ is a homotopy from $p\tilde f=f$ to $p(x_0)$ meaning that $f$ is null-homotopic.

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Thank you $SL_2$! – Jiangnan Yu May 9 '12 at 23:42

I'll assume by $P^2$ you mean the real projective plane and I'll also assume you know that $\pi_1(P^2)=\mathbb{Z}/2$.

Now, every map $f:(X,x_0) \to (Y,y_0)$ of based topological spaces induces a map between their fundamental groups $f_*:\pi_1(X,x_0) \to \pi_1(Y,y_0)$.

Now, $\pi_1(P^2,x_0)=\mathbb{Z}/2$ (and in fact, we may forget about the base point, since $P^2$ is path connected). Also, $\pi_1(S^1, x_0)=\mathbb{Z}$. So a continous map $f:P^2 \to S^1$ induces, by the above paragraph, a map $\mathbb{Z}/2 \to \mathbb{Z}$. Torsion elements are sent to torsion elements, but $\mathbb{Z}$ is torsion-free, so the only possibility is that $\mathbb{Z}/2$ is mapped to zero. So every map $f:P^2 \to S^1$ induces the trivial map on fundamental groups. This means that $f_*$ is homotopic to the null map. EDIT: See SL2's answer below for the conclusion.

(regarding maps $\mathbb{Z}/2 \to \mathbb{Z}$: What happens to $1 \in \mathbb{Z}/2$? The map must satisfy $f(1+1)=f(1)+f(1)$, but $f(1+1)=f(0)=0$ , so $2f(1)=0$, hence $f(1)=0$.)

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You also need that the higher homotopy groups of $S^1$ vanish. Normally it is not sufficient to check $\pi_1$. (Equivalently, $f$ lifts to a map to the universal cover, which is contractible in this case.( – Grumpy Parsnip May 9 '12 at 16:18
@Jim Thank you for pointing that out. I found this relevant MO post which says that even if $f_*\colon \pi_k(X)\to \pi_k(Y)$ is 0 for all $k$, it isn't true that $f$ is necessarily null-homotopic. You really do need a covering space argument to finish this. – SL2 May 9 '12 at 16:33
@SL2: And thank you for pointing that out. – Grumpy Parsnip May 9 '12 at 17:07
@JimConant : Thanks for pointing out my error. I've edited my answer accordingly (with a reference to the answer below). – Fredrik Meyer May 9 '12 at 19:53
Thanks Fredrik Meyer,now I know $\mathbb Z/2 \to \mathbb Z$could only be a trivial map :) – Jiangnan Yu May 9 '12 at 23:43