Finite fundamental group and covering spaces

Question

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?

Answer 1

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.

Answer 2

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 $Y$.