# Fundamental group of the $T^1 \cup D^2$

Let $T^1 =\mathbb{S}^1 \times \mathbb{S}^1$ be the torus, $D^2$ the unit disk and $X = T^1 \cup D^2$. Find the fundamental group of $X$.

Try: I believe there should be a simpler argument applying Seifert Van Kampen's Theorem, but I was unable to find $U, V$ open such that $X = U \cup V$ in the first place, so I tried looking at the space $X$ as the adjunction space obtained by gluing the disk to the torus along the unit circle $\{ x^2 + y^2 = 0, z = 0 \}$. Define a map $f : \mathbb{S}^1 \to T^1$ by $$e^{2\pi i t} \mapsto (e^{2 \pi i t}, 1)$$ Then we get a commutative diagram ($i$ and $\pi_i, i =1,2$ denote the inclusion and canonical maps to the quotient space)

$\require{AMScd}$ \begin{CD} \mathbb{S}^1 @>{i}>> D^2\\ @V f\ VV @V \pi_1\ VV\\ \mathbb{S}^1 \times \mathbb{S}^1 @>{\pi_2}>> (\mathbb{S}^1 \times \mathbb{S}^2) \cup_f D^2 \end{CD} which induces a commutative diagram on the fundamental groups

\begin{CD} \pi_1(\mathbb{S}^1) @>{i_*}>> \pi_1(D^2)\\ @V f_*\ VV @V \pi_{1*}\ VV\\ \pi_1(\mathbb{S}^1 \times \mathbb{S}^1) @>{\pi_{2*}}>> \pi_1((\mathbb{S}^1 \times \mathbb{S}^2) \cup_f D^2) \end{CD} Now, since $\pi_{2*} \circ f_* = \pi_{1*} \circ i_*$, take $\gamma(t) = (\cos t, \sin t)$ on $\mathbb{S}^1$, then $$(\pi_{1*} \circ i_*)([\gamma]) = \pi_{1*}([i \circ \gamma]) = \pi_{1*}([\gamma]) = [\epsilon]$$ since $D^2$ is simply connected, where $\epsilon$ denotes the trivial loop. On the other hand $$(\pi_{2*} \circ f_*)([\gamma]) =\pi_{2*}([f \circ \gamma]) = \pi_{2*}([\sigma]) = [\epsilon]$$ and by the way we defined $f$, $[\sigma] = [f \circ \gamma]$ is a generator of the torus. So by attaching the disk to the torus we have "killed off" one generator of the torus, and thus the fundamental group of $X$ is $$\pi_1(X) = \{0 \}\times \mathbb{Z} \cong \mathbb{Z}$$

Is this proof okay? Thank you in advance.

• How is the disc attached to $D^2$ to form $X$? In your attempt you take $X$ to be the cofiber of the inclusion $i_1:S^1\hookrightarrow S^1\times S^1$, which has the homotopy type of $S^1\vee S^2$, whose fundamental group is calculable via a covering space argument. – Tyrone Nov 20 '17 at 15:38
• What did you mean by "disk attached to $D^2$? The torus is obtained by rotating the circle $(x-2)^2+z^2=1$ along the $z$ axis, so $T^1\cup D^2$ is a space such that $T^1\cap D^2 = S^1$. And I didn't understand the rest of your comment, I don't know about "cofibres". If you could expand that idea a bit more I'd be grateful. – user313212 Nov 20 '17 at 15:48
• Typo. Meant "How is the disc attached to $T^1$ to form $X$". The torus is most often obtained as a quotient of $I\times I$ by suitably identifying edges, but I understand your notation now. – Tyrone Nov 20 '17 at 17:37
• Try seeing it this way. Attach your 2-disc to the torus by filling the central hole. By symmetry of $T^1=S^1\times S^1$ this is the same as the other possible way of attaching $D^2$. Now the disc is contractible, and contracting it to a point inside $X$ gives you what looks like a torus but with no hole in the middle: what was the inside circle now meets at a single point (draw this). We have a quotient map $X=T^1\cup D^2\xrightarrow{\simeq}X/D^2$ which is a homotopy equivalence. – Tyrone Nov 20 '17 at 17:45
• Now consider $S^2\cup I$, where we attach a line running from the North to the south pole inside the 2-sphere. Contracting this line is a homotopy equivalence which gives you the same object as you got by contracting the disc in $X$ (convince yourself of this). That is $S^2\cup I\xrightarrow{\simeq}(S^2\cup I)/I\simeq X/D^2$. On the other hand we can keep the line and slide its end points along the surface of $S^2$ till they meet, without changing the homotopy type. The end result is a wedge $S^2\vee S^1$. Thus we have $X\simeq X/D^2\simeq (S^2\cup I)/I\simeq S^2\cup I\simeq S^2\vee S^1$. – Tyrone Nov 20 '17 at 17:48