# projective space, constant function, homotopy

Show, that every continious function $f:\mathbb{R}P^2\to S^1$ is a homotopy to a constant function.

(Hint: Has $f$ a lift $\tilde{f}:\mathbb{R}P^2\to\mathbb{R}$)

Hello,

I want to solve this problem, but I am kinda stuck and the given hint just confuses me even more... I would appreciate, if someone could clarify this a bit.

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Edit:

I want to solve this by a way Eduard Longa gave:

Let $p:\mathbb{R}→S^1$ be the standard covering map. Since $\mathbb{R}$ is contractible, the map $f$ is homotopic to a constant function via an homotopy $H$. Then, the composition $p\circ H$ provides an homotopy between $f$ and a constant function.

I want to go through this, step by step. First of all I want to show, that $\mathbb{R}$ is contractible.

Therefore I have to show, that $\mathbb{R}$ is homotopic to a set $\{x_0\}$.

$f_1:\mathbb{R}\to\{x_0\}$, $f_1(x)=x_0$

$f_2:\{x_0\}\to\mathbb{R}$, $f_2(x)=x$

I have to show, that $f_1\circ f_2\sim id_{\{x_0\}}$ and $f_2\circ f_1\sim id_{\mathbb{R}}$

$(f_1\circ f_2)(x_0)=x_0$

$(f_2\circ f_1)(x)=x_0$

To show, that $f_1\circ f_2\sim id_{\{x_0\}}$ I give the homotopy $H_1:\{x_0\}\times[0,1]\to\{x_0\}$ simply by $H_1(x,t)=x_0$ and for $f_2\circ f_1\sim id_\mathbb{R}$ similar $H_2:\mathbb{R}\times [0,1]\to\mathbb{R}$, with $H_2(x,t)=(1-t)x+tx_0$.

Hence $\mathbb{R}$ is contractible. (It is completly trivial) Am I right?

The next step is to show, that $f$ is homotopic to a constant function via an homotopy $H$. Is here $\tilde{f}$ meant? Since $f:\mathbb{R}P^2\to S^1$ I do not know, why $\mathbb{R}$ contractible, has the consequence, that $f$ is homotopic to a constant function. Respectively how this observation about $\mathbb{R}$ helps.

• Do you understand why $f$ has a lift? – Eduardo Longa Jun 19 '16 at 1:23
• Actually I never asked myself that question. So no, I do not think so. :( – MrTopology Jun 19 '16 at 1:24
• You can have a look at the section "Lifting properties" at this page of wiki en.m.wikipedia.org/wiki/Covering_space . Since the fundamental group of the projective plane has order 2, the homomorphism induced by $f$ in fundamental group is the trivial one. Hence, by the lifting property in that article of wiki guarantees the existence of a lifting to $\mathbb{R}$, the universal cover of $S^1$. – Eduardo Longa Jun 19 '16 at 1:28
• Is there an easy way to see, that the fundamental group of the projective plane has order 2? We have not showed that yet. – MrTopology Jun 19 '16 at 1:37
• Easy, I'm not sure...There is a properly discontinuous action of $Z_2$ in the sphere $S^2$ by means of the antipodal map $p \mapsto -p$. The quotient space of this action is the projective space. Then, by a theorem, the fundamental group of the quotient space is isomorphic to the group acting on the space (sphere, in this case), since the sphere is simply connected. – Eduardo Longa Jun 19 '16 at 1:43

In general if $X$ is contractible then every continuos map $f:X\rightarrow Y$ is contractible. A brief explication: if you have $f_1 :X\rightarrow pt$ be the constant map and $f_2 :pt\rightarrow X$ such that $f_1 \circ f_2 \sim Id_{pt}$ and $f_2 \circ f_1 \sim Id_{X}$ then clearly :

$f\sim f\circ Id_X\sim f\circ f_2 \circ f_1$ but $f_1$ is the constant map so $f\circ f_2 \circ f_1$ is constant too and you have realized your homotopy.

Now you have to prove the existence of the lifting $\tilde{f}$. As Eduardo Longa pointed out you can use the fact that the fundamental group of the projective plane has order 2. If you don't want to calculate it, I suggest the following alternative:

You have the quotient map $q:S^2\rightarrow \mathbb{RP}^2$ induced by the antipodal equivalence, so you have the map $g:=f\circ q$ from $S^2$ to $S^1$. Since $\pi_1(S^2)=0$ you can lift $g$ in the following way (as pointed out by Eduardo Longa) with $p$ the exponential map.

\begin{matrix} S^2&\stackrel{\tilde{g}}{\longrightarrow}&\mathbb{R}\\ &\stackrel{g}{\searrow}&\downarrow{p}\\ &&S^1 \end{matrix}

Now Let's prove that $\tilde{g}$ induces a lifting $\tilde{f}$: consider the map $\phi:S^2\rightarrow \mathbb{R}$ defined as $\phi(x)= \tilde{g}(x)-\tilde{g}(-x)$, then it is a continuous map with integer values (since $p(\tilde{g}(x))=g(x)=g(-x)=p(\tilde{g}(-x)))$ so the image is a set of points, but $S^2$ is connected so it's just a point. So we have for all $x$ $\phi(x)=\phi(-x)$ but we have also $\phi(x)=-\phi(-x)$ so we obtain that $\phi$ is the zero map.So $\tilde{g}(x)=\tilde{g}(-x)$ and we can descend it to a lifting $\tilde{f}$

You conclude using the fact that $p$ is contractible (since $\mathbb{R}$ is contractible).

Let me put the whole argument together; hopefully you can fill in the gaps.

Let $f:\mathbb{RP}^2 \to S^1$ be any continuous map. It induces a map on fundamental groups (after suitable choices of basepoint, which I omit) $f_\star:\pi_1(\mathbb{RP}^2) \to \pi_1(S^1)$. But the only such homomorhism is the zero map. (For this, you need to know what each fundamental group is.) By the lifting criterion (Proposition 1.33 in Hatcher), $f$ has a lift $\widetilde{f}: \mathbb{RP}^2 \to \mathbb{R}$ (that is, there exists an $\widetilde{f}$ such that $p \circ \widetilde{f}=f$ for $p:\mathbb{R}\to S^1$ the standard covering map). Now, as noted above, $\widetilde{f}$ must be homotopic to a constant map since $\mathbb{R}$ is contractible. Let $F$ be a homotopy from $\widetilde{f}$ to a constant map. Then $p \circ F$ is a homotopy from $f$ to a constant map.