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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!

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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.

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  • $\begingroup$ Instead you can just observe that the Klein bottle group maps to Z with kernel isomorphic to Z. $\endgroup$ – Moishe Kohan Sep 20 '14 at 20:24
  • $\begingroup$ 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? $\endgroup$ – Nephry Sep 20 '14 at 20:25
  • $\begingroup$ @studiosus I'm not sure I understand what you're trying to say or how it's relevant. $\endgroup$ – Adam Hughes Sep 20 '14 at 20:26
  • $\begingroup$ @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? $\endgroup$ – Adam Hughes Sep 20 '14 at 20:27
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    $\begingroup$ @AdamHughes Haha, I'm sorry, I think I need a cup of coffee, it's absolutely clear now; thank you very much, again! $\endgroup$ – Nephry Sep 20 '14 at 20:31

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