Hatcher Exercise 1.2.8 via the van Kampen theorem

Question

Hatcher's Exercise 1.2.8 is the following : Compute the fundamental group of the space $X$ obtained from two tori $S^1 \times S^1$ by identifying a circle $S^1 \times \{x_0\}$ in one torus with the corresponding circle $S^1 \times \{x_0\}$ in the other torus.

There are solutions on this site already, as in the discussion here, where the original poster mentions that one needs open sets in order to apply the van Kampen theorem. I'd like to follow up on this thought with the solution I found here (and also shown below).

Here are my questions about the provided solution:

• What allows us to take an open neighborhood of $S^1 \times \{x_0\}$ that deformation retracts onto $S^1 \times \{x_0\}$? Does any space admit such an open neighborhood that deformation retracts onto it? Or is there something special about $S^1 \times \{x_0\}$ that admits this open neighborhood?
• Applying the van Kampen theorem here requires $X$ to be the union of two path-connected open sets with path-connected intersection. How is $X$ being written as a union of two path-connected open sets here? Is $X = (T_1 \cup U_1) \cup (T_2 \cup U_2)$? If so, how can we see that $T_1 \cup U_1$ and $T_2 \cup U_2$ are path-connected and open, and have path-connected intersection?
• Below is the statement of the van Kampen theorem in Hatcher. With this in mind, it looks like the normal subgroup $N$ in the provided solution is $\pi_1(S^1)$. How can one see that this is the case?

Thanks!

Answer 1

• In the more general case of two copies $A \times B$ joined by identifying the two copies of $A \times \{x_0\}$ for some $x_0 \in B$, as long as $x_0$ has a neighborhood $U$ in $B$ that deformation retracts to $\{x_0\}$, then $A \times U$ is a neighborhood of $A \times \{x_0\}$ that deformation retracts to $A \times \{x_0\}$. Every point of $S^1$ has neighborhoods isomophic to intervals on the real line. Intervals are contractible.
• There is a slip-up on the indexing in this part of the proof. Since $U_1 \subset T_1, T_1 \cup U_1 = T_1$ and similarly $T_2 \cup U_2 = T_2$. What they meant was the two open sets are $T_1 \cup U_2, T_2 \cup U_1$. Let $C = T_1 \cap T_2$ be the common circle. $T_1 \cup U_2$ is open since it is the image under the quotient of the open set $T_1 \sqcup U_2$ in $T_1 \sqcup T_2$. Similarly for $T_2\cup U_1$. Since $U_2$ retracts to $C$, $T_1 \cup U_2$ retracts to $T_1 \cup C = T_1$ and $U_1 \cup U_2$ retracts to $U_1 \cup C = U_1$, which itself retracts to $C$. Similarly $T_2 \cup U_1$ retracts to $T_2$. Since $C \cong S^1$, it is path-connected, and $T_1, T_2$ are path-connected. Since $U_1 \cup U_2$ retracts to $C$, it is also path-connected.
• The retractions above induce group isomorphisms $$\pi_1(T_1 \cup U_2) \cong \pi_1(T_1)\\\pi_1(T_2 \cup U_1) \cong \pi_1(T_2)\\\pi_1(U_1 \cup U_2) \cong \pi_1(C) \cong \pi_1(S^1)$$