How do I prove this statement about trivial covering spaces? I have the following problem about covering spaces, but first of all I write our definitions:

Definiton isomorphism of coverings An isomorphism between covering maps $p:Y\rightarrow X$ and $p':Y'\rightarrow X'$ is a homeomorphism $\phi:Y\rightarrow Y'$ such that $p'\circ \phi=p$


Definition of a trivial covering We say that a covering $p:Y\rightarrow X$ is trivial if $\forall x\in X$ we can take $X\in \mathfrak{U}(x)$ s.t. $p^{-1}(X)=\dot\bigcup_\alpha N_\alpha$ where

*

*$N_\alpha$ are open forall $\alpha$

*$N_\alpha\cap N_\beta =\emptyset$ forall $\alpha \neq \beta$

*$p|_{N_\alpha}:N_\alpha \rightarrow X$ us a homeomorphism


Now I have the following statement:

A covering $p:Y\rightarrow X$ is trivial iff it is isomorphic to a covering of the form $$p':X\times Z\rightarrow X;\,\,(x,z)\mapsto x$$ Where $Z$ is any set with the discrete topology.

I wanted to prove this statement, but I somehow struggle a bit.
My Idea was the following. Consider $$\phi:Y\rightarrow X\times Z;\,\,\,y\mapsto (p(y),z)$$ Then clearly $p'\circ\phi(y)=p'(p(y),z)=p(y)$ for all $y\in Y$ but now I remark that for $\phi$ to be homeomorphic it needs to be homeomorphic in each component. At the moment it is only true for the first component since in the second component it is not continuous. I think if I can find another second component I'm done right?
Could someone help me finding this second component?
Thanks
 A: The idea is relatively easy, in fact $Y=p^{-1}(X)=\cup_{\alpha\in I}U_\alpha$ and you have to prove that there exists a covering isomorphism between $X\times I$ and $Y$, where there is the discrete topology on $I$.
The map $f: Y\to X\times I$ is of course the following one
$f(y)=(p(y),\alpha)$ if $y\in U_\alpha$
$f$ is well defined because of the definition of trivial covering.
Moreover $f$ is continuos if and only if $j_1\circ f: Y\to X$ and $j_2\circ f: Y\to I$ are continuos maps, where $j_i$ are the natural projections of the product $X\times I$.
You can observe that $p=j_1\circ f$ so that $j_1\circ f$ is continuos.
To prove that $j_2\circ f$ is continuos you have to prove that the inverse image of a point $\alpha \in I$ is an open set of $Y$. By definition you have that
$(j_2\circ f)^{-1}(\alpha)=U_\alpha$ so that $j_2\circ f$ has to be continuos.
Thus $f$ is a continuos map.
Now we can define the inverse map
$g: X\times I\to Y$ sending $(x, \alpha)$ to $y$ such that $y\in U_\alpha\cap p^{-1}(x)$. This map is well defined by the definition of trivial covering.
If $U$ is an open set of $Y$, then
$g^{-1}(U)=\sqcup_\alpha g^{-1}(U_\alpha\cap U)=\sqcup_\alpha p(U)\times \{\alpha\}$
that is an open set because $p$ is an open map, and so $p(U)$ is an open set of $X$ and $I$ has the discrete topology.
Now you have that $g$ is a continuos map and it’s the inverse map of $f$, thus $f$ is an homeomorphism.
Now you can define $p’: X\times I\to X$ as the projection map on the first factor. Of course this map is a covering map and you can verify by your hands that
$p=p’\circ f$
Please observe that $I$ has an important meaning. In fact corresponds to the inverse image of a certain point $x \in X$
$I\cong p^{-1}(x)$, for each $x \in  X$
Where the topology on $p^{-1}(x)$ is the induced topology of $Y$. Please prove that this topology is exactly the discrete topology on $p^{-1}(x)$ by definition of trivial covering.
This means that a trivial covering is in general a disjoint union of $p^{-1}(x)$- copies of $X$.
