1

We have a simply connected space $X$ and the mapping torus $M_f \cong (X \times [0,1])/p $ where $p$ identifies $(x,0)$ with $(f(x), 1)$. For example: how do we compute the fundamental group of $X \times [0, \frac{1}{2}]$ under the image? if $f(X)$ is again simply connected then the fundamental is clearly trivial, but this may not be the case.

Now since $M_f$ is path connected, we consider w.l.o.g the basepoint $x_0 = p(x,1/2) \in M_f$ where $x \in X$.

My plan was to consider $X_1, X_2$ the images of $X \times [0, \frac{1}{2} + \epsilon], X \times [\frac{1}{2}, 1] \cup C$ under $M_f$ respectively, where $C$ denotes a circle basically connecting $p(x,0)$ and $p(x, 1/2)$, hence $X_0 = X_1 \cap X_2$ is again connected and $x_0 \in X_0$.

So we use Seifert-van-Kampen and get the following diagram:

$$\require{AMScd} \begin{CD} \pi_1(X_0, x_0) @>{}>> \pi_1(X_1, x_0)\\ @V{}VV @VV{}V\\ \pi_1(X_2, x_0) @>>{}> \pi_1(X, x0)\end{CD}$$

Now we only have to calculate the respective fundamental groups, but I don't know how to proceed. Any feedback would be appreciated, I've tried looking for similar questions on the website: Details for calculating the fundamental group of mapping torus and Fundamental group of mapping torus?. But they seem to be either skipping this detail or using some complicated tools that are not readily available for me yet.

hteica
  • 319
  • Which of the terms and maps in that square have you been able to calculate? – ronno Jun 03 '23 at 11:56
  • @ronno sadly I have problem with calculating the fundamental groups since they contain either $X \times {0}$ or $X \times {1}$ under the image of $M_f$. As for the maps, they are all inclusions so just identities. – hteica Jun 03 '23 at 11:59
  • @ronno how to see that $X_1 \simeq X$ and $X_2 \simeq f(X) \vee S^1 $? – hteica Jun 03 '23 at 12:00
  • Do you know why the mapping cylinder of $X \to Y$ is homotopy equivalent to $Y$? Use this to show that $X_2 \simeq X \vee S^1$ and $X_0 \simeq X \vee X \vee S^1$. $X_1 \simeq X$ because it deformation retracts to $X \times 0$. – ronno Jun 03 '23 at 12:07
  • @ronno but at the same time we have $X \times 0 = f(X) \times 1$ under the image of $M_f$ does this not affect the fundamental group? – hteica Jun 03 '23 at 12:11
  • There is something to show here but essentially, because there are no identifications on $X \times [0,1)$, it turns out that $X \times [0, 1)$ maps homeomorphically onto its image in $M_f$. – ronno Jun 03 '23 at 12:14
  • @ronno yes that part is clear, but what happens with $X \times {0}$? there's an identification there. – hteica Jun 03 '23 at 12:31
  • @ronno apologies, I meant, there's an identification between $(x, 0)$ and $(f(x), 1)$ – hteica Jun 03 '23 at 12:38

0 Answers0