Fundamental group of a quotient of a cylinder 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?

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