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