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


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.

Thanks in advance.



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.

  • 3
    $\begingroup$ Do you understand why $f$ has a lift? $\endgroup$ – Eduardo Longa Jun 19 '16 at 1:23
  • $\begingroup$ Actually I never asked myself that question. So no, I do not think so. :( $\endgroup$ – MrTopology Jun 19 '16 at 1:24
  • 5
    $\begingroup$ 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 $S^1$. $\endgroup$ – Eduardo Longa Jun 19 '16 at 1:28
  • $\begingroup$ Is there an easy way to see, that the fundamental group of the projective plane has order 2? We have not showed that yet. $\endgroup$ – MrTopology Jun 19 '16 at 1:37
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.