# For the space $X = T \# T$ find its fundamental group and prove the universal cover is contractible.

Let $X = T \# T$, the sum of $2$ tori identified by removing a small disk from each of them and gluing the missing disks along the boundary. Find the fundamental group of $X$, show the universal cover is contractible, and deduce that any image of $S^n$ into $X$ is contractible.

As for the first part i assume one should apply Van Kampen to the spaces of the Tori with a disk removed (so that the intersection is homotopically equivalent to the $S^1$ around the removed part). Now I think the torus with a disk removed is homotopically equivalent to a torus with a point removed, which is the wedge of two circles. But one of this circles is generated by the intersection in Van Kampen, so that one we consider to be a trivial loop. In the end Van Kampen gives us that $X$ has the fundamental group of $Z * Z$ which coincedentally is again the fundamental group of a wedge of two circles.

Now my understanding of Van Kampen is dubious at best, is this line of thinking correct? And how do i go about finding the universal cover? There's a hint to the second part of the task to look at covers of $X$ homotopically equivalent to a wedge product of circles, but I'm not sure how to even define such covers. How would one proceed with the rest of the task?

As pointed out in the comments by multiple people:

The original line of thinking about the fundamental group makes no sense and comes from a misinterpreted van Kampen. The proper answer given by the theorem is the group $<a,b,x,y| [a,b] = [x,y]>$.

What remains is the part about the universal cover

• This sentence: "But one of this circles is generated by the intersection in Van Kampen, so that one we consider to be a trivial loop." is false, so the rest of your argument is wrong. If the two natural generators of the torus (the meridian and longitude) are called $a$ and $b$, then the boundary of the gluing circle is homotopic to $aba^{-1}b^{-1}$. – John Hughes Aug 31 '17 at 0:22
• @JohnHughes Could you help clarify this for me: what you say makes sense when i look at the diagram of the torus, but since the fundamental group of the torus is abelian shouldnt the generators commute making $aba^{-1}b^{-1}$ trivial? – WardenOfTheMath Aug 31 '17 at 0:33
• In the sum construction, you've cut out part of the torus that is needed to make the generators of the fundamental group commute. – Rob Arthan Aug 31 '17 at 0:48
• @RobArthan Right. So we cut out the part that lets the generators commute, but Van Kampen says that $aba^{-1}b^{-1}$ is trivialized in the final group anyway? So is the answer to the question of fundamental group $(Z \times Z ) * (Z \times Z)$? – WardenOfTheMath Aug 31 '17 at 0:53
• You have two spaces with fundamental group free on two generators (these are the punctured tori). The identification (based on @johnhughes comment) identifies the commutator in one of these free groups with the commuator in the other (here we're talking about commutators of generators). The fundamental group is then $\langle a,b,x,y\mid [a,b]=[x,y]\rangle$ by van Kampen. – Steve D Aug 31 '17 at 0:59

The universal cover of $T \# T$ is a simply connected surface. It cannot be the sphere, so it has to be non-compact, and so its second homology is 0. Hence clearly all of its homologies are zero, and since it's simply connected, it's contractible by Whitehead's theorem.