Fundamental group of two tori with a circle ($S^1✕${$x_0$}) identified

Question

Compute the fundamental group of the space obtained from two tori $S^1✕S^1$ by identifying a circle $S^1✕${$x_0$} in one torus with the corresponding circle $S^1✕${$x_0$} in the other.

Using van Kampen's theorem, I can fairly quickly show that if $T_1$ is one torus and $T_2$ is the other, then the group has to be isomorphic to $((π_1(T_1) ∗ π_1(T_2)))/N ≅ (\mathbb{Z} ✕ \mathbb{Z})∗ (\mathbb{Z} ✕ \mathbb{Z})/N$, where N is generated by elements of the form $i_{12}(w)i_{21}(w)^{-1}$, w is in $\pi_1(S^1✕${$x_0$}$) ≅ \mathbb{Z}$ and $i_{12}$, $i_{21}$ are the maps from $\pi_1(S^1✕${$x_0$}$)$ to $π_1(T_1)$, $π_1(T_2)$ respectively determined by inclusion. My problem is, as it seems to always be with van Kampen-related problems, figuring out what the elements of N look like.

My attempt to do this in a pure-algebra way, as far as I can tell, failed me: I tried examining the case where w represents a loop that goes n (some integer) times around the circle, but then it seems like $i_{12}(w)$ "$=$" $(n, 0)$ in $T_1$ and $i_{21}(w)$ "$=$" $(n, 0)$ in $T_2$, which says to me that $i_{12}(w)i_{21}(w)^{-1} $"$=$"$ (n, 0) + (-n, 0) = 0$. The notation here isn't right, since $(n, 0)$ in $T_1$ is not the same as $(n, 0)$ in $T_2$ and their addition (composition?) shouldn't be written quite that way, hence my quotes around the equals signs, but certainly no loop in $S^1✕${$x_0$} is going to magically include anything but the 0 loop in either torus's second component, right? But this doesn't seem right to me, since if a, b are the generators of the circles in $T_1$ and d, e are the generators of the circles in $T_2$, and c is the constant path then the identification of the circle should make, say, $a = d$, so for example $(a, b)*(d, e)$ $ = (a^2,b)*(c, e)$ $ = (c, b)*(d^2,e)$. This doesn't match with what I just worked out N might be, but I also can't figure out what N should look like to make this equivalence make sense.

My attempt to understand this geometrically has gone even worse. What I know for sure is that the resulting figure isn't the "2-torus", 2 tori glued together at a disc. What I don't know for sure is whether this identification should actually create the hypertorus $S^1✕S^1✕S^1$ or something completely different.

Answer 1

Here are two tori with a copy of $S^1$ identified. The picture is clearer when we move $x_0$ to the inner rim of one torus, and to the outer rim on the other torus.

The two images below show a path on one of the tori that belongs to the homotopy class of the latter generator.

Basically Van Kampen says that these two become homotopic in $\pi_1$ of the combination. Geometrically this is obvious, as you can slide that path from one torus to the other via the shared circle of contact.

More precisely, if $b$ denotes the homotopy class above and $a_1$ (resp. $a_2$) is a loop around the tube of the bigger (resp. smaller) torus, then elements of the amalgamated product are objects like $$ (a_1^{k_1},b^{\ell_1})*(a_2^{m_1},b^{n_1})*\cdots*(a_1^{k_t},b^{\ell_t})*(a_2^{m_t},b^{n_t}), $$ where the exponents $k_i,\ell_i,m_i,n_i$ are all arbitrary integers, $t$ is a natural number and factors with $a_1$ (resp. $a_2$) alternate.

Because $b$ commutes with both $a_1$ and $a_2$, the above product simplifies to $$ a_1^{k_1}a_2^{m_1}a_1^{k_2}a_2^{m_2}\cdots a_1^{k_t}a_2^{m_t}b^r $$ with $$ r=\sum_{i=1}^t(\ell_i+n_i). $$ This is path that first does $k_1$ laps around the outer tube, then $m_1$ laps around the inner loop et cetera, and at some point also goes $r$ laps along the intersection.

Looks like $(\Bbb{Z}*\Bbb{Z})\oplus\Bbb{Z}$ as a group. A direct product of two groups where the first factor is a free product of two infinite cyclic groups, and the latter is just $\Bbb{Z}$.

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In retrospect this is clear. Undoubtedly you know that the fundamental group of figure 8 is that free product of two infinite cyclic groups. Your space is the direct product $8\times S^1$.

Answer 2

I'll add that this means the group is isomorphic to $\mathbb{Z} \times F$, where $F$ is a free (nonabelian) group on two generators.

In particular, before modding out, there are two pairs of generators, say $\{x,y\}$ for the first $\mathbb{Z}\times \mathbb{Z}$ and $\{a,b\}$ for the other. Each pair commutes with itself, but the two pairs do not commute. Then, modding out by $N$ just says $y = b$.

So now we have three generators, $y(=b), x, a$. The first one commutes with both the other two, and $x$ and $a$ don't commute. That gives $\mathbb{Z} \times F$.