In the proof of Van Kampen theorem using the covering space theory, we are using that if $p_1 : E_1 \rightarrow U_1$ and $p_2 : E_2 \rightarrow U_2$ are two covering space, such that $B=U_1 \cup U_2$, $U_0 = U_1 \cap U_2$ and $\psi : p_1^{-1}(U_0) \rightarrow p_2^{-1}(U_0)$ is an isomorphism of covering space (for the covering space $p_{{i}_{|p_i(U_0)^{-1}}} : p_i^{-1}(U_0) \rightarrow U_0$ induced by $p_i$ for $i = 1,2$), then if we note $X= (E_1 \sqcup E_2) / <\psi(x_1) \sim x_1 \quad \forall x_1 \in p_1^{-1}(U_0)>$
we have then, if we consider : $p:E_1 \sqcup E_2 \rightarrow B$ which send $x_1 \in U_1$ to $p_1(x_1)$ and $x_2 \in U_2$ to $p_2(x_2)$, then $p$ is well defined and constant on the equivalence class, and then : $\overline{p} : X \rightarrow B$ is well defined, continuous and is a covering space.
Actually, the point which is not clear and I try to write it, is that $\overline{p}$ is a covering space. How to prove it ?
Thank you !
Edit : Actually, after an other try, I did something. Let $x \in B = U_1 \cup U_2$. We can suppose $x \in U_1$. Let $V_1$ be an open set which is an evenly covered neighborhoods for $p_1$, $x \in V_1$. If $V_1 \cap U_0 = \emptyset$, then we consider $W = V_1$ (case 1). If $V_1 \cap U_0 \neq \emptyset$, let $V_2$ be an open set which is an evenly covered neighborhoods for $p_2$, with $V_2 \subset V_1 \cap U_0$, $x \in V_2$. In this last case, we note $W = V_1 \cap V_2 = V_2$. We have then : $\overline{p}^{-1}(W) = \overline{p}^{-1}(V_2) = q(\bigsqcup_{i \in I} W_{1,i} \cup \bigsqcup_{j \in j} W_{2,j})$, with : $$ p_1^{-1}(W) = \bigsqcup_{i \in I} W_{1,i}$$ $$ p_2^{-1}(W) = \bigsqcup_{j \in J} W_{2,j}$$
Moreover, $V_2 \subset V_1 \cap U_0$, then we have :
$p_i^{-1}(W) = p_{{i}_{|p_i(U_0)^{-1}}}^{-1}(W)$. But $\psi$ is an isomorphism of covering space, so we have :
$p_1^{-1}(W) = \psi^{-1} \circ p_2^{-1}(W) = \bigsqcup_{j \in J} \psi^{-1}(W_{2,j})$
and then :
$\overline{p}^{-1}(W) = q(\bigsqcup_{j \in J} W_{2,j}) = \bigsqcup_{j \in J} q(W_{2,j})$, which seems to give us what we want.
In the case 1, it seems that we can just use the fact that $p_1$ is a covering space of $U_1$, cause here there is no "interfence" between $U_1$ and $U_2$.
It is the right idea ?
I think that you have the good idea. If you are in case 1 there is no interference and the proof is what we see on a drawing. If you are in the case where $x\in U_1\cap U_2$ it is quite similar. But if you are in the case $x\in \overline{U_0}-U_0$ it gets more tricky and I think that you forgot this case, because if you assume the existence of your open $V_2$ you have $x\in V_2\subset V_1\cap U_0\subset U_0$ which is not always the case if $U_0\cap V_1\not=\emptyset$.
Edit: Here is a picture for case 1
Edit 2: I think there is nothing much to change in the proof. We write $$p^{-1}(V_1) = \bigsqcup_j W_j$$ and $W_j^\prime =\psi(W_j\cap p_1^{-1}(U_0))$. Then $$\overline{p}^{-1}(V_1)=\bigsqcup_j (W_j\cup W_j^\prime)/_\psi$$
There is just to prove that $$\overline{p}:(W_j\cup W_j^\prime)/_\psi\longrightarrow V_1$$ is a homeomorphism which I think can be done easily using the factorization theorem.
Question: Do you have a reference for the complete proof of Van Kampen using covering spaces please?