The mapping torus \(T_f\) of a map \(f: X \to X\) is the quotient of \(X \times I\) obtained by identifying each point \((x, 0)\) with \((f(x), 1)\). In the case \(X = S^1 \lor S^1\) with \(f\) basepoint-preserving, compute a presentation for \(\pi_1(T_f)\) in terms of the induced map \(f_*: \pi_1(X) \to \pi_1(X)\). Do the same when \(X = S^1 \times S^1\). [One way to do this is to regard \(T_f\) as built from \(X \lor S^1\) by attaching cells.]
Solution
Let \(x_0\) be the wedge point of \(X\) and assume \(f(x_0)=x_0\). Let \(t\) denote the loop in the mapping torus traced by the basepoint as the \(I\)-coordinate runs once around the quotient.
1. If \(X=S^1\vee S^1\) with free generators \(a,b\) of\[ \pi_1(X,x_0)\cong F(a,b), \]then \(T_f\) is obtained from\[ X\vee S^1 \]by attaching two \(2\)-cells.
Give each circle of \(X\) one \(0\)-cell and one \(1\)-cell. The product of each \(1\)-cell with \(I\) descends in the mapping torus to a \(2\)-cell. The \(1\)-skeleton of the mapping torus consists of the two original loops \(a,b\) together with the loop \(t\) coming from \(x_0\times I\). Thus its \(1\)-skeleton is \(X\vee S^1\).
2. The two attaching maps impose\[ t a t^{-1}=f_*(a), \qquad t b t^{-1}=f_*(b). \]
Use the copy \(X\times\{1\}\) to name the generators \(a\) and \(b\), and orient \(t\) from level \(0\) to level \(1\) along the basepoint. Consider the square obtained from the cylinder over the \(a\)-edge. Its oriented boundary traverses the bottom edge, the right vertical edge, the top edge backwards, and the left vertical edge backwards. Because the quotient identifies \((x,0)\) with \((f(x),1)\), the bottom edge represents \(f_*(a)\) in the chosen top copy, while the reversed top edge represents \(a^{-1}\). Hence the boundary word is \[ f_*(a)\,t\,a^{-1}\,t^{-1}, \] so its relation is equivalent to \[ t a t^{-1} f_*(a)^{-1}=1. \] The same argument for the \(b\)-edge gives the second relation.
3. Therefore\[ \boxed{ \pi_1(T_f) \cong \left\langle a,b,t\ \middle|\ t a t^{-1}=f_*(a),\ t b t^{-1}=f_*(b)\right\rangle . } \]
The \(1\)-skeleton contributes the free group on \(a,b,t\), and 2 gives exactly the attaching relations of the two \(2\)-cells. Van Kampen for CW complexes yields the displayed presentation.
4. Now suppose\[ X=S^1\times S^1 \]with the standard generators \(a,b\) of\[ \pi_1(X)\cong\mathbb Z^2 \cong\langle a,b\mid [a,b]=1\rangle. \]Then\[ \boxed{ \pi_1(T_f) \cong \left\langle a,b,t\ \middle|\ [a,b]=1,\ t a t^{-1}=f_*(a),\ t b t^{-1}=f_*(b) \right\rangle . } \]
2.The product of the torus \(2\)-cell with \(I\) gives a \(3\)-cell and therefore introduces no additional relation in the fundamental-group presentation.
5. The two requested presentations are therefore those in 3 and
4.
These are exactly the two cases stated in the source exercise.