# Map from $\mathbb{R}\mathbb{P}^{2}$ to Klein bottle homotopic to constant map

I'm currently struggling with the following exercise:

Show that any continuous map

$$f: \mathbb{R}\mathbb{P}^{2} \to K$$

where $K$ is the Klein bottle, is homotopic to a constant map.

I know that $\pi_{1}(K,k_{0})$ has presentation

$$\pi_1(K,x_0)\cong \langle a,b | aba^{-1}b\rangle$$

and $\pi_{1}(\mathbb{R}\mathbb{P}^{2},x_{0}) \cong \mathbb{Z}_{2}$ is a group of order $2$, so my first attempt was to show that $\pi_{1}(K,k_{0})$ does not contain an element of order $2$, which would imply that the image of the induced map

$$f^{*}:\pi_{1}(\mathbb{R}\mathbb{P}^{2},x_{0})\to \pi_{1}(K,k_{0})$$

is trivial and hence, by the lifting theorem, $f$ lifts to the universal cover of $K$, which would show the claim.

Unfortunately though, and I might just be missing something silly, that proved to be rather tricky. I just don't really know where to go from there.

Any thoughts or different ideas to prove it much appreciated!

You are right that it is sufficient to see if there are any elements of order $2$ in $\pi_1(K,x_0)$.

Since $ab=b^{-1}a$ and you can write an arbitrary word in the free group $\Bbb Z *\Bbb Z$ as

$$a^{e_0}b^{f_1}a^{e_1}b^{f_2}\ldots a^{e_n}b^{f_{n+1}}$$

where $e_0, f_{n+1}\in\Bbb Z$ and $e_i, f_i\in\Bbb Z\setminus\{0\}$ for $1\le i\le n$.

We see that we can move $b$s past $a$s in our group until all the $b$s are on the left and all the $a$s on the right. For example, we reduce the word $ab^2a^3b$ as follows:

$$ab^2a^3b=abbaaab$$ $$=(ab)(baaab)$$ $$=b^{-1}(ab)(aaab)$$ $$=b^{-2}aaa(ab)$$ $$=...$$ $$=b^{-6}a^4$$

and, more generally, when we see a block like $a^mb^\ell$ we see that we move each $b$ across $m$ different $a$s, so that this is seen to mean

$$a^mb^{\ell}=b^{-m\ell}a^m$$

we see that any word in $\pi_1(K,x_0)$ is equivalent to some $b^na^k$ with $n,k\in\Bbb Z$.

From here, we assume that this element is of order $2$, i.e.

$$b^na^kb^na^k=e$$

by reducing the word on the LHS, we get

$$b^{n}b^{-kn}a^{2k}=e\implies n-kn=0\iff k=1$$

From this we conclude that our condition is just

$$b^nab^na=a^2=e\iff a^2=e$$

but $a$ is a generator with no torsion restriction in the presentation, and so has infinite order.

• Instead you can just observe that the Klein bottle group maps to Z with kernel isomorphic to Z. – Moishe Kohan Sep 20 '14 at 20:24
• Thank you very much for your answer, I should have thought of actually reducing the words! The only thing I don't see is how the original relation also imposes the relation a(b^-k) = (b^-(k+1))a or am I just missing the forest for trees right now? – Nephry Sep 20 '14 at 20:25
• @studiosus I'm not sure I understand what you're trying to say or how it's relevant. – Adam Hughes Sep 20 '14 at 20:26
• @Nephry that's not a relation I'm claiming, I'm saying if you have $ab=b^{-1}a$ then $a^mb^\ell=b^{-m\ell}a^m$. Where do you see this other one? – Adam Hughes Sep 20 '14 at 20:27
• @AdamHughes Haha, I'm sorry, I think I need a cup of coffee, it's absolutely clear now; thank you very much, again! – Nephry Sep 20 '14 at 20:31