$f:X\rightarrow S^1$ a continuous map. $X$ a path-connected topological space.

Let $f:X\rightarrow S^1$ be a continuous map from a path-connected topological space $X$ and let $p:\mathbb{R} \rightarrow S^1$ be the universal covering.

What is the condition when $f$ admits a lifting that there exists a continuous map $h: X \rightarrow \mathbb{R}$ such that $p \circ h = f$ ?

• Have you looked in any textbook which treats covering spaces? Jun 18 '14 at 8:13
• @MarianoSuárez-Alvarez I have read some notes but not textbook Jun 18 '14 at 8:16
• This is a condition on the pushforwards of the fundamental groups: See on the "Lifting Conditions" part: en.wikipedia.org/wiki/Covering_space Jun 18 '14 at 8:29
• Note that there are no non trivial homomorphisms from a finite group to the infinite cyclic group, therefore the image on the level of fundamental group of such spaces will always be trivial , ie contained in the image of $p_*$ Jun 18 '14 at 8:52

In general the following result holds for all connectet, locally connected topological spaces:

Let $p:X\longrightarrow Y$ be a cover and let $f:Z\longrightarrow Y$ be a continuous map. Let $z_{0}\in Z$ be a fixed basepoint and let $y_{0}=f(z_{0})\in Y,$ $x_{0}\in X$ such that $p(x_{0})=y_{0}$ (i.e., all the base point are constructed in the obvious way).

Hence, the continuous map $f$ admits a lift if and only if the following condition is satisfied:

$$f_{\ast}\pi_{1}(Z,z_{0})\subseteq p_{\ast}\pi_{1}(X,x_{0})$$

In particular, if $Z$ is simply connectd, your map always admits a lift. Moreover, here's an application to your specific case. First of all, you know $\pi_{1}(S^{1},y_{0})=\mathbb{Z}.$ As a useful criterion, if $\pi_{1}(Z,z_{0})$ is an abelian group of the form $(\mathbb{Z}/p_{1}^{q_{1}}\mathbb{Z})\times\cdots\times(\mathbb{Z}/p_{r}^{q_{r}}\mathbb{Z}),$ where $p_{1},\ldots,p_{r}$ are prime numbers, then you can deduce the group homomorphism $f_{\ast}$ must be $f_{\ast}=0$ and so your map $f$ admits a lift. This is, for example, the case of $Z=\mathbb{RP}^{n}$ for $n\geq2.$

• And in this case $X$ is given as the universal cover, so we need the pushforward/induced map to be zero (the zero class) in Y. I don't know how to generalize this; I guess the image of any class must not contain a full loop about $S^1$. Jun 18 '14 at 8:44
• In your specific case, $\pi_{1}(S^{1},y_{0})=\mathbb{Z}.$ As a useful criterion, if $\pi_{1}(Z,z_{0})$ is an abelian group of the form $\mathbb{Z}/p^{\alpha_{1}}\mathbb{Z}\times\mathbb{Z}/p^{\alpha_{1}}\mathbb{Z}$ Jun 18 '14 at 8:58
• That is beyond my level of expertise in Algebra at this moment. That is a good point, why don't you include your comment in your post? Jun 18 '14 at 9:09
• Sorry, but I make a mystake in writing my previous comments: I'm new here. Here the application to your case and an example. In your specific case, $\pi_{1}(S^{1},y_{0})=\mathbb{Z}.$ As a useful criterion, if $\pi_{1}(Z,z_{0})$ is an abelian group of the form $(\mathbb{Z}/p_{1}^{q_{1}}\mathbb{Z})\times\cdots\times(\mathbb{Z}/p_{r}^{q_{r}}\mathbb{Z}),$ where $p_{1},\ldots,p_{r}$ are prime numbers, then you can deduce the group homomorphism $f_{\ast}$ must be $f_{\ast}=0$ and so your map $f$ admits a lift. This is, for example, the case of $Z=\mathbb{RP}^{n}$ for $n\geq2.$ Jun 18 '14 at 9:11
• This is a good point; include it in your post and I will upvote you; you can just copy and paste your comment into the post. I learnt this interesting issue from your comment. Welcome/Wilkommen to Math Stack Exchange. Jun 18 '14 at 9:14

If you can show that every possible homomorphism between a finite group and $\mathbb Z$ "factors through", you are done.I think the issue here is that there is only one possible homomorphism between a finite group and $\mathbb Z$, so that, in particular, there is only one possible induced homomorphism between the two fundamental groups.

Figure out what that homomorphisms is/are and show that that the homomorphism(s) satisfies the lifting conditions. Notice that a homomorphism must map elements of finite order to elements of finite order. Are there (non-trivial) elements of finite order in $\mathbb Z$? If not so, what is/are the only possible homomorphisms between the two groups?