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Let $X$ be a connected topological space which admits a universal covering. Let $Y$ be the topological space obtained as following:we have an homeomorphism $f : X \to X$ , we take $X \times \left[0,1\right]$ and we identify $(x,0)$ with $(f(x),1)$.

I should prove that $\pi_1(Y)$ is a semidirect product of $\pi_1(X)$ and $\mathbb{Z}$.

I think I found a solution but the fact that does not convince me is that I did not use the universal covering hypothesis.

Let $$\pi: X \times \mathbb{R} \to Y$$ $$(x,t) \to (f^{-\left \lfloor{t}\right \rfloor}(x),\{t \}) .$$ This map is a covering map.

One can check that the automorphism group is $\mathbb{Z}$ and acts transitively on the fiber of this covering; actually, one can explicitly describe this group as the maps $$\alpha_n(x,t)=(f^n(x),t+n)$$ with $n$ an integer.

So, because of standard theory facts about normal covering one has $$\frac{\pi_1(Y)}{\pi_1(X)} \cong \mathbb{Z} .$$

Now, a reasoning of group theory should end the demonstration.

Where is the mistake and where should I have used that hypothesis?

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  • $\begingroup$ Please provide a proof that the map you wrote is a covering map. $\endgroup$ – Steve D Sep 2 '18 at 13:11
  • $\begingroup$ It is not clear to me even why $\pi$ is continuous. $\endgroup$ – Pedro Tamaroff Sep 2 '18 at 14:59
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Note that the homeomorphism $f$ gives an isomorphism $f_*=\pi_1(f) : \pi_1(X) \longrightarrow \pi_1(X)$, so that $\mathbb Z$ acts on $\pi_1(X)$ via $n\mapsto f_*^n$. In this way you obtain a group $G = \pi_1(X) \rtimes \mathbb Z$. Now recall that if $\pi : \tilde{X}\to X$ is the universal cover of $X$, there is an isomorphism between the group of deck transformations of $\pi$ and $\pi_1(X)$. I will produce from this universal cover of $X$ a cover of $Y = IX/((x,0) = (f(x),1))$.

Since $f$ is a homeomorphism, it lifts to a homeomorphism $\tilde f$ of $\tilde{X}$ such that $\pi f =\tilde{f} \pi$. You can now consider the disjoint union $\tilde Y$ of the spaces $\tilde X[i] \times [0,1]$ with $i\in\mathbb Z$ where $(\tilde x_i,0)$ identifies with $(\tilde f(\tilde x_{i+1}),1)$. There is a (well defined) map $\rho : \tilde Y\to Y$ that sends $[\tilde x_i,t]$ to $[\pi\tilde x_i,t]$ and this is the universal cover of $Y$.

It suffices then to show that the group $D(\rho)$ of deck transformations of $\rho$ is a semidirect product of $D(\pi)$ and $\mathbb Z$. The definition of $\rho$ shows that at least the group $D(\rho)$ contains a copy of $D(\pi)$, where you can make any deck transformation of $\pi$ act on the first coordinate; you also have a copy of $\mathbb Z$ acting by shifting through $\tilde f$. Perhaps you can take it from here and conclude.

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  • $\begingroup$ There are some details missing, notably showing your universal cover is simply connected. A possibly easier approach is to build your covering space out of $X$ instead; this space is homotopy equivalent to $X$, and then getting the fundamental group structure (and even the semidirect action) is easier. $\endgroup$ – Steve D Sep 2 '18 at 16:07
  • $\begingroup$ @SteveD I would be glad to have this answer improved or have a better answer posted. I am no serious algebraic topologist, so this is the best I could come up with. :) $\endgroup$ – Pedro Tamaroff Sep 2 '18 at 16:09
  • $\begingroup$ No your approach is perfect! I was simply mentioning that the group structure falls out a bit better if you don't pass to the universal cover first. Showing the space is simply connected is not completely trivial, but I think that's fine to leave to the reader :). If I have time later, I'll post an answer as well. $\endgroup$ – Steve D Sep 2 '18 at 16:12
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Your idea is sound, but the map you wrote down is not a covering map (at least it's not obviously one to me).

The space you're talking about is the mapping torus, and we can cover it with something called the mapping telescope.

Let $Y_\mathbb{Z}=\sqcup (X\times[n,n+1])$, and let $\widehat{Y}$ be the quotient space of $Y_\mathbb{Z}$ identifying $(x,n+1)$ and $(f(x),n+1)$, where the former is a right endpoint, and the latter a left endpoint. We can further write $\widehat{Y_m}$ as the image of $\{(x,n)\mid |n|\le m\}$ in $\widehat{Y}$.

Now $\widehat{Y}$ is a covering space of your $Y$, with covering map $\rho([x,n])=[x,n\pmod{1}]$. The sheets of this covering look like $$ \rho^{-1}([x,t])=\{[f^m(x),t+m]\mid m\in\mathbb{Z}\}$$

Now the deck transformations act on $\widehat{Y}$ via $p_m([x,n])=[f^m(x),n+m]$. Since this action is transitive on fibers, $\rho_*\pi_1(\widehat{Y})$ is a normal subgroup of $\pi_1(Y)$, and we also have $\pi_1(Y)/\rho_*\pi_1(\widehat{Y})\cong\mathbb{Z}$, with $\mathbb{Z}$ corresponding to those deck transformations. Since $\mathbb{Z}$ is a free group, we actually have a split extension $$ \pi_1(Y)\cong\pi_1(\widehat{Y})\rtimes\mathbb{Z}$$

Finally, we have $\pi_1(\widehat{Y})\cong\pi_1(X)$. That follows by describing an explicit homotopy equivalence between $X$ and $\widehat{Y_m}$ [for any $m$], and then reasoning via compactness that any map $\alpha:S^1\rightarrow\widehat{Y}$ ends up in some $\widehat{Y_m}$. Note that we can actually describe the semidirect action as well, since we have the deck transformations above that generate the $\mathbb{Z}$ factor. Namely, for $[\alpha]\in\pi_1(X)$ and $m\in\mathbb{Z}$, the action is given by $$ [\alpha]^{(m)}=[f^m(\alpha)]$$

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Someone suggested to me the following alternative approach. Pick a basepoint $\ast\in X$ and consider the covering map $g : Y\to S^1$ that sends $[x,t]$ to $\exp(2 \pi i t)$. This induces a map $g_*:\pi_1(Y) \to \pi_1(S^1) = \mathbb Z$. By picking a loop from the $\ast$ to $f(\ast)$, you can construct a splitting for $g_*$. This means that $\pi_1(Y)$ is the semidirect product of $\mathbb Z$ with the kernel of $g_*$, so it suffices to show that $\ker g_*$ is $\pi_1(X)$. The fibre of $g$ at the basepoint of $S^1$ is equal to the space obtained by glueing $X$ to $X$ along $f$, and since $f$ is an homeomorphism, this is just $X$. Since $\pi_2(S^1)=0$, you obtain from the LES of a fibration what you wanted, a (split) exact sequence

$$0\to \pi_1(X) \to \pi_1(Y)\to \mathbb Z\to 0.$$

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