Disk glued to a Torus. Classifying the resulting spaces – How/Where to begin

Question

My algebraic topology professor gave me an exercise and I feel very lost.
The exercise is the next:

Describe the homotopy type of the different spaces that can be obtained from gluing a disk to a torus.

Obs: The gluing is only at the disk's border.

On a first try, I was thinking of gluing the border of the disc to a single side of the torus (>), then to two sides (maybe continuous sides (>,>>) or inverting the orientation of the first gluing(>,<)), etc.

After a week, I gave up and asked my prof for hints. He said that could be useful to think in $\mathbb{R}^2/\mathbb{Z}^2$ and he also said that at some point I would be working on an algebra problem instead of a topology problem.
To do the classification of the spaces, he also mentioned that
Van Kampen would be necessary.

Any ideas/suggestions about this?

Answer 1

Hint: The attaching map $\phi:S^1\to T^2$ is a continuous map, meaning that if $\gamma$ generates $\pi_1(S^1)$, then $\phi_*(\gamma)$ represents a class of loops $[\phi_*(\gamma)]\in\pi_1(T^2)$, where $\phi_*:\pi_1(S^1)\to\pi_1(T^2)$ is the induced homomorphism. Now, you need to invoke Seifert-Van Kampen's Thm and the fact that the fundamental group is topologically invariant.

(My solution is in the spoiler.)

Let $X=T^2\cup_\phi D^2$. Take $A=B^2\subset D^2,\ B=(X\setminus B^2)^\epsilon$ ($B^2$ is an open ball contained in the attached disk), where the supscript $\epsilon$ means "a small open neighborhood of", and the two nhbds deformation retracts onto $\{*\}, T^2$ respectively. It is easy to see that by a proper choice of neighborhoods, we have $A\cap B\simeq S^1$. Now, we apply Seifert-Van Kampen's Thm, which gives $$\pi_1(X)\cong(\pi_1(A)*\pi_1(B))/{\langle[\phi_*(\gamma)]\rangle}\cong\pi_1(B)/{\langle[\phi_*(\gamma)]\rangle}\cong\pi_1(T^2)/{\langle[\phi_*(\gamma)]\rangle}$$ Recall that the attaching map $\phi$ induces a homomorphism $\phi_*:\Bbb Z\to\Bbb Z^2$ on the fundamental groups. This map is completely determined by $\phi_*(1)=(a,b)$, so \begin{align}\pi_1(X)\cong\Bbb Z^2/{\langle(a,b)\rangle}\cong\Bbb Z\times\Bbb Z_{\gcd(a,b)}&\tag{1}\end{align} This tells us that $X_{a,b}\not\simeq X_{a',b'}$ if $\gcd(a,b)\neq\gcd(a',b')$. It's worth noting that isomorphic fundamental groups does not give you the same homotopy type. It seems that $X_{0,3}\not\simeq X_{3,0}$ despite the fact that $\pi_1(X_{0,3})\cong\pi_1(X_{3,0})$.

Proof of $(1)$: Using Bezout's identity, we get that $d=\gcd(a,b)=ua+vb$ for some $u,v\in\Bbb Z$. We define a surjective homomorphism \begin{align}&f:\Bbb Z\times\Bbb Z\to\Bbb Z_d\times\Bbb Z\\ &f(x,y)=\left(ux+vy+d\Bbb Z,\frac{a}{d}y-\frac{b}{d}x\right)\end{align} We know that $\ker f=\langle(a,b)\rangle$, so the result follows from the first isomorphism theorem.