find the map which induces the homorphism between fundamental groups

Denote $X=\mathbb{R}P^2$ to be the 2-dimensional real projective space, see definition here wikipedia.

Also from the above link, we know that it has a CW complex structure with one 0-cell $x_0$, one 1-cell and one 2-cell.

Then suppose we have a group homomorphism $\phi: \pi_1(X,x_0)\rightarrow \pi_1(Y,y_0)$, where $Y$ is a path-connected space, prove that we can find a map $f:(X,x_0)\rightarrow (Y,y_0)$, such that $f_*=\phi.$

Remark: this is a special case of one homework, I know that we can first define a map $f:S^1\rightarrow Y$ to be $f(a)=b,$ where $\phi([a])=[b]$ and $S^1$ is the 1-cell, $[a]$ is the generator of $\pi_1(X)=\mathbb{Z}/{2\mathbb{Z}}$, but how to extend it to the whole $X$, i.e., to the whole 2-cell?

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Key lemma for inductively defining mappings out of cell complexes: if the map $f_n$ is defined on the $n$-skeleton $X_n$ already, and you want to define it on an $n+1$-cell with attaching map $\phi: S^n \to X_n$, then you can do so if and only if $f_n \phi$ is nullhomotopic.

So you need to show that the attaching map from the circle to the 1-skeleton of $\mathbb{RP}^1$ becomes nullhomotopic after applying the map $f$ you've constructed so far, but that just says that $2[a]$ needs to be 0 on $Y$, which is true.

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I have not learnt this key lemma, could you give me some reference for this? –  ougao Oct 26 '12 at 2:24
Hatcher talks about it at the bottom of p. 90. Jason DeVito's answer is the proof of the lemma in this case (and his argument is what works in general). –  user29743 Oct 26 '12 at 2:37
Oh, I find that, thank you! –  ougao Oct 26 '12 at 3:11

Since $[b]$ is the image of $[a]$, and $[a]$ is 2-torision, $[b]$ must be 2-torsion. That means that $2b$ is null homotopic. Let $F:S^1\times I$ be such a null homotopy. For definiteness, assume $F(x,0) = y_0$ and $F(x,1) = 2b$.

Since $F(x,0)$ is constant, $F$ descends to a map on $S^1\times I/$~ where we identify all points of the form $(x,0)$. It's easy to see that $S^1\times I/$~ is homeomorphic to a disc $D$.

So, we have a map $\tilde{F}:D\rightarrow Y$ which maps the boundary circle of $D$ to $2b$. It follows that $\tilde{F}$ descends to a map on $D/$~ where we now identify the antipodal points on the boundary of $D$. But $D$ with these points identified is homeomorphic to $\mathbb{R}P^2$, we we finally get our map $\tilde{\tilde{F}}:\mathbb{R}P^2\rightarrow Y$. This is the map you want.

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This is essentially a proof of countinghaus's key lemma in a very special case. –  Jason DeVito Oct 26 '12 at 2:25
Thank you! got it. –  ougao Oct 26 '12 at 3:12