Fundamental group of Mapping Torus $M_h$: how to prove that the action is really $h_*$?

Question

Let us work with the following setting:

let $h$ be an automorphism (assume base point preserving) of a genus $g$ surface ($g>0$) to himself. $h \colon (\Sigma^g,\ast) \to (\Sigma^g,\ast)$ and define the mapping torus $M_h$ in the usual way.

Notice that $h$ induces an action $\pi_1(S^1,\ast)\to Aut(\pi_1(\Sigma^g,\ast))$ just by sending the chosen generator to $h_*$, so it is indeed plausible the existence of the semidirect product: $\pi_1(\Sigma^g,\ast)\rtimes_{h_*} \pi_1(S^1,\ast)$.

I was asked to prove that the fundamental group $\pi_1(M_h, \ast)\cong \pi_1(\Sigma^g,\ast)\rtimes_{h_*} \pi_1(S^1,\ast)$ (with a little abuse of notation one can identify all the basepoints to a chosen one).

By the l.e.s. of the fibration $\Sigma^g \to M_h \to S^1$ it's easy to see that $$\pi_1(M_h, \ast)\cong \pi_1(\Sigma^g,\ast)\rtimes_{?} \pi_1(S^1,\ast)$$

Being $\pi_1(S^1,\ast)\cong \mathbb{Z}$, $\pi_2(S^1,\ast)\cong 0$ and on the zero level the inclusion of the fibre induces a bijection, we have the following s.e.s. $$ 0 \to \pi_1(\Sigma^g,\ast) \xrightarrow{incl.} \pi_1(M_h, \ast) \xrightarrow{\pi} \pi_1(S^1,\ast)\to 0$$ which is right-split (since $\mathbb{Z}$ is free).

Please notice the "?" I've put: I don't have the slightest idea on how to determine that the action there is really the one induced by $h$, because my reasoning above was purely algebraic. I know abstractly the action is given by conjugation, but how to prove the this action is the same as the one induced by $h$?. My attempt was to try and find something from that l.e.s. but I don't see anything helpful there. for what concern S.v K. I can't find an helpful covering of the mapping torus.

I'm aware that there is another analogue question here, but it is $5$ years old and the answer doesn't provide any insights, nor the comments.

Answer 1

The mapping torus is a union of two copies of $\Sigma_g\times\left[0,1\right]$ where the gluing of one boundary component is via identity and of the other boundary component is via $h$.

One can not apply Seifert-van Kampen directly to this decomposition because the intersection is not connected.

Therefore one uses the following trick. Denote $M_h=X_1\cup X_2$ the above decomposition. In a first step consider the union $Y_1=X_1\cup C$, where $C$ is a circle which is obtained as a union of $\left\{x_0\right\}\times\left[0,1\right]\subset \Sigma_g\times\left[0,1\right]=X_2$ with a suitable interval in $X_1$. The intersection $X_1\cap C$ is just $\left\{x_0\right\}\times\left[0,1\right]$, in particular it is connected and you can apply Seifert-van Kampen which gives $\pi_1Y_1=\pi_1X_1* {\mathbb Z}$.

Now $M_h$ is the union of $Y_1$ and $X_2$ and their intersection is connected because it is obtained by joining the two copies of $\Sigma_g$ by an arc between them. So you can now apply Seifert-van Kampen to this union to compute $\pi_1M_h$.

Answer 2

Here are two ways to see this for a general mapping torus $M_f$ and $f:N\to N$:

1) Bass Serre Theory: take open neighborhoods $U,V$ of $N\times [0,\frac 12]$ and $N\times[ \frac 12,1]$. The fundamental group of $M_f$ computes as fundamental group of the graph of groups of this open cover. The graph of groups is given by two vertices $U$ and $V$ connected with two edges coming from the two components of $U\cap V$. Note that the groups of all vertices and edges are $\pi_1N$ and that all but one edge inclusion are the identity. A maximal tree contains only one edge (for computation preferrably the one with the two identity morphisms), which immediately gives the fundamental group (by slight abuse of notation) as $$ \langle \pi_1N,t|txt^{-1}= f(x)\rangle, $$ where $t$ comes as generator from the remaining edge.

2) The other way to see this is to regard the action of the fundamental group of the base space on the fiber (exists for all Serre fibrations). It is easy to compute that this action is precisely the same as the action coming from $0 \to \pi_1N \to \pi_1M_f\to \pi_1S^1\to 0$. Once the action is coming from topology and once from algebra. However the one coming from topology is on the nose :-)

2*) For regular $G$-covering spaces we have the fiber sequence $0 \to \pi_1\tilde M \to \pi_1M\to G\to 0$ (note that $G$ is this time $\pi_0$ of the fiber!). You can check that the Deck group action of $G$ coincides the the $G$-action coming from the exact sequence. Hence by inserting the correct cover $G=\mathbb Z$, $\tilde M=N\times \mathbb R$, $M=M_f$ (the pullback of the universal $\mathbb Z$-cover along the fibration) and the deck group action on $N\times \mathbb R$ is one the nose once again.

Bonus question: what is the relationship between 2) and 2*)?