# Finite fundamental group and covering spaces

Show that if a path-connected, locally path-connected space X has a finite fundamental group , then every map $X$ to $S^1 \times S^1$ is nullhomotopic (i.e. homotopic to a constant map) .

Is the same true if we replace the torus with the wedge sum of two circles?

I was able to solve the first part using that the fundamental group of the covering space: $R \times R$ is trivial, but regarding the second part, about the wedge sum of two circles, I think the issue is that the fundamental group of the covering space of the wedge sum is not trivial,is my intuition true? or I am missing something?

• 5 months being a member and and 16 questions asked? About time to learn the easy rules to properly write mathematics in this site: meta.math.stackexchange.com/questions/5020/… – Timbuc Apr 10 '15 at 13:03
• also thank you :) I will next time @Pedro – Butterfly Apr 10 '15 at 13:09
• Ok let me explain what I have done and why. I first placed the question you found somewhere in a book or course in one block. After that, I made a new block with your personal question regarding to the problem. Then I made a last block with all the attempts you made. Then it is immediatly clear for everyone: (a) what the original question was you needed to solve, (b) what your personal question is, and (c) what you tried to solve that personal question. :) – Pedro Apr 10 '15 at 13:15
• @Pedro thank you I really appreciate your help – Butterfly Apr 10 '15 at 13:26
• If by trivial you mean simply-connected, then note that there always exists a trivial covering space of $X$ when $X$ is path-connected, locally path-connected and semi-locally simply-connected. In particular, $S^1\vee S^1$ has a simply-connected covering space, which is presented at page 59 of Hatcher's book. – Stefan Hamcke Apr 10 '15 at 13:37

basically it is true for any space $Y$ with fundamental group $G$ s.t there is no non-trivial homomorphism from $\pi_1(X) \to G$...and universal cover of $Y$ is contractible (basically $Y$ is a $K(G,1)$ space)...then by using map lifting criterion you can always get a lifting of your function $f$ on the universal cover where the image is contractible and composition with covering map will give you a null homotopy map for $f$ in $Y$.

• Can I suppose that X is R since the fundamental group of R is trivial hence finite, and then take f to be the functions that gives me a loop on one of the circles of the wedge sum, then this loop can't be shrunk to a cte loop ,therefore f is not nullhomotopic? is this true ? @jesusRs – Butterfly Apr 17 '15 at 11:54
• Since R is contractible so any continuous map is null homotopic.... Think!! – Anubhav Mukherjee Apr 17 '15 at 14:14

Any mapping $f:X\to S^1\times S^1$ induces a homomophism $f_*:G=\pi(X)\to\pi(S^1\times S^1)=\mathbb Z\times\mathbb Z$. If $G$ is finite, so is $f_*(G)$, and since the only finite subgroup of $\mathbb Z\times\mathbb Z$ is $\{0\}$, we conclude $f_*(G)=\{0\}$ and by the lifting criterion there is $\widetilde f:X\to\mathbb R\times\mathbb R$. Clearly $\widetilde f$ is nullhomotopic, hence so is $f$.

Now, let $X=S^1\wedge S^1$ be the wedge of two circles, say $X=S^1\cup C$ where $C$ is another circle tangent to $S^1$ at a point $p$. You can extend the identity $S^1\to S^1$ to $g:X\to S^1$, just collapsing $C$ onto $p$. Finally compose $g$ with $S^1\equiv S^1\times\{p\}\subset S^1\times S^1$ to get $f:X\to S^1\times S^1$. We claim that $f$ is not nullhomotopic.

To see that, consider the fundamental group $\pi_1(X,(p,p))=\mathbb Z\times\mathbb Z$. This fundamental group is generated by the two fundamental loops $\sigma=S^1\times\{p\}$ and $\tau=\{p\}\times S^1$. Thus, the homomorphism $f_*$ induced in homotopy groups has the fundamental loop $\sigma$ in the image, hence that image cannot be trivial. So $f_*\ne0$ and consequently $f$ cannot be nullhomotopic.

• Excuse me but i have a question About the second part , i don't understand the part where we map the circle to a loop and we get it in the image of f* can you explain it to me or write it down please ? I'm really thankful for your help @JesusRS – Butterfly Apr 13 '15 at 20:02
• I've rewritten a bit the post. – Jesus RS Apr 13 '15 at 21:45
• @Butterfly: It seems the answer is not accepted. But the important thing is whether there is something unclear and I can help you to solve the problem. – Jesus RS Apr 15 '15 at 18:48
• The thing is that I read both answers and I got lost because they are somehow contradicting each other , and regarding your answer , I don't get the S^1 x {p} what is p and why f* of the loop we didn't collapse is indeed this S^1 x{p} ? I'm really sorry for taking too long to understand it .. Thank you so much :) @JesusRS – Butterfly Apr 15 '15 at 19:09
• I've tried to explain better my answer. Concerning the other answer, it explains the general facts concerning mappings $X\to S^1\times S^1$ where $X$ has finite fundamental group. The key property is that for such an $X$ all homomorphisms $\pi_1(X)\to G=\pi_1(S^1\times S^1)$ are trivial. – Jesus RS Apr 15 '15 at 20:10