I am working on this problem:

The mapping torus $T_f$ of a map $f : X → X$ is the quotient of $X×I$ obtained by identifying each point $(x,0)$ with $(f(x),1)$. In the case $X = S_1∨S_1$ with $f$ basepoint-preserving, compute a presentation for $π_1(T_f)$ in terms of the induced map $f_∗ : π_1(X) → π_1(X)$. Do the same when $X = S_1 ×S_1$.

There are 3 different versions of solutions that I find: version 1:

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version 2: enter image description here

version 3: enter image description here

Now I am confused. By different methods they gives different conclusions. I think all of them sounds reasonable. So which is correct? Or are they equivalent? Why?

Thanks for any help!

  • $\begingroup$ The last one is correct, and in general this is what happens. The interval that closes to a circle in the gluing becomes a new generator $g$, and there are new 2-cells with boundary $g^{-1}xgf_\ast(x)$ for original generators $x$. $\endgroup$
    – Steve D
    Oct 13, 2020 at 20:07

1 Answer 1



I was wrong. The last one is right by tracing the attaching $1-$cells. You can verify that by doing the case when $X = S^1\wedge S^1$ with $f=id$ easily, which you have probably done in question $1.2.8.$ In that case, the fundamental group should be $\mathbb{Z}\oplus(\mathbb{Z} * \mathbb{Z})$, which is given by the last solution, while with the first and second solution, the answer is $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$.

My Original Answer:

In comparison with the work that I have done with this problem, this first solution is right. As for the second one, for the case when $X = S^1 \wedge S^1$, the answer is the same as the first solution. The author did not do the case when $X= S^1\times S^1$. According to the third solution, there were some kind of conjugates involved, which I think might be a mistake due to messing up with the base-point, but I am not a hundred percent sure, since it is kind of hard to prove two group presentations are different. I hope someone could help with this.


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