Let $p : E → B$ and $p' : E' → B$ be two covering spaces. Show that the fiber product $E \times_B E'$ is a covering of $B$.
So I know the categorical definition of fiber product. And I think in this case the product induces two projection maps $\pi:E_1 \times_B E_2\rightarrow E_1$ and $\pi':E_1 \times_B E_2\rightarrow E_2$ such that $p_1\circ \pi=p_2\circ \pi'$. But I don't really know how to proceed. My feeling is that we can somehow use the categorical language to prove this, without resorting to the basic definition of a covering space, but I'm not sure how. Any hint is appreciated!
Hi all, I've drafted the proof and decided to post it here so you people can help to judge if it's okay. Your help is appreciated! (Sorry that the proof is very verbose, but I just want to make sure I understand how to juggle with the definitions)
Note that $E \times_B E'$ also has a natural map $q$ to $B$, sending $(e,e')$ to $p(e) = p'(e')$. We want to show that $q: E\times_B E'\rightarrow B$ is a covering space. For any $b\in B$, since $p : E \to B$ and $p' : E' \to B$ are covering spaces, there exists open neighborhoods $U_B$ and $U_B'$ of $b$ such that $p^{-1}(U_B)\overset{h}{\cong} U_B \times \Gamma$ and $p'^{-1}(U_B') \overset{h'}{\cong} U_B' \times \Gamma'$. Take $U=U_B\cap U_B'$ then $p^{-1}(U)\cong U \times \Gamma$ and $p'^{-1}(U)\cong U \times \Gamma'$. We claim that $q^{-1}(U)\overset{g}{\cong} U\times(\Gamma \times \Gamma')$, where we define $g(e,e')=(p(e),(p_2\circ h(e), p_2\circ h'(e')))$ and $p_2$ is the projection of product onto the second coordinate.
Given any $(e_1,e_1'),(e_2,e_2')\in q^{-1}(U)$, then by definition of $q$, we have $e_1,e_2\in p^{-1}(U)$ and $e_1',e_2'\in p'^{-1}(U)$. If $g(e_1,e_1')=g(e_2,e_2')$ by construction of $g$, we have $p(e_1)=p(e_2)$, $p'(e_1')=p'(e_2')$ $p_2\circ h(e_1)=p_2\circ h(e_2)$, and $p_2\circ h'(e_1'))=p_2\circ h'(e_2'))$, which together implies that $e_1=e_2$ and $e_1'=e_2'$ so $g$ is injective. We observe that $g$ is also surjective because $h$ and $h'$ are surjective, i.e., given $(b^*,(m,n))\in U\times(\Gamma \times \Gamma')$, we can let $e=h^{-1}(b^*,m)$ and $e'=h'^{-1}(b^*,n)$, then $g(e,e')=(b^*,(m,n))$. Thus, $g$ is a bijection.
The continuity of $g$ follows from composing continuous functions $p,p_2,h$ and $h'$ gives continuous function. For $g^{-1}$, we can explicitly construct it by defining $g^{-1}(b^*,(m,n))=(h^{-1}(b^*,m), h'^{-1}(b^*,n))$. It's straightforward to verify this is indeed the inverse of $g$. Again, since $h,h'$ are homeomorphisms, $g^{-1}$ is a composition of continuous functions and therefore continuous.
Thus, we had obtained the desired homeomorphism $g$. Since $\Gamma$ and $\Gamma'$ are discrete, $\Gamma\times \Gamma'$ is also discrete. Since $p$ and $p'$ are covering maps, $q$ is obviously surjective. Finally, we will show that $q$ is continuous. Given any open set $U\in B$, it's evenly covered by $p$ and $p'$ so $p^{-1}(U)\times p'^{-1}(U)$ is a collection of open sets in $E\times E'$. Because $E\times_B E'$ has topology induced by inclusion in $E\times E$, $q^{-1}(U)\subset p^{-1}(U)\times p'^{-1}(U)$ is open in $E\times_B E'$ so $q$ is indeed continuous.
Therefore, we conclude that $q:E\times_B E'\rightarrow B$ is indeed a covering space.