Are the connected components of the pullback of a connected covering space all isomorphic? Given a connected covering space $Y\rightarrow X$ (both $Y$ and $X$ are connected). Let $T\rightarrow X$ be any map, with $T$ connected. The pullback $Y\times_X T$ may not be connected.
At least in the case of $Y/X$ Galois, the pullback will be a disjoint union of isomorphic covers of $T$ (the Galois group acts transitively on the connected components). Without the assumption that $Y/X$ is Galois, is it still true that the connected components of $Y\times_X T$ are isomorphic as covers of $T$?
I'm happy to assume that $T\rightarrow X$ is a finite covering map, and $Y\rightarrow X$ is also finite, if that helps.
 A: If $T$ is locally path-connected and your covering map $p:Y\to X$ is regular, then answer is yes. In this case $T\times_X Y$ is locally path-connected and thus the restriction of the pullback covering map $q:T\times_XY\to T$ to each (open) path component is a covering map. 
Call $f:T\to X$ your map. Since $T$ is path connected, each path component of $T\times_XY$ meets every fiber of $q$. So to use the classification of covering maps, we may consider path components $P_1$ and $P_2$ of $T\times_XY$ and points $(t,y)\in P_1$ and $(t,y')\in P_2$ (notice $p(y)=f(t)=p(y')$). Using the universal property of the pullback you can show that $q_{\#}(\pi_1(P_1,(t,y))=f_{\#}^{-1}(p_{\#}(\pi_1(Y,y)))$ and $q_{\#}(\pi_1(P_2,(t,y'))=f_{\#}^{-1}(p_{\#}(\pi_1(Y,y')))$. Since $p$ is regular, $p_{\#}(\pi_1(Y,y))=p_{\#}(\pi_1(Y,y'))$ and so $q_{\#}(\pi_1(P_1,(t,y))=q_{\#}(\pi_1(P_2,(t,y'))$. The classification of covering maps then tells you that $q:P_1\to T$ and $q:P_2\to T$ are equivalent, and in particular, that $P_1$ and $P_2$ are homeomorphic. If you knew pullbacks of conjugate subgroups in $\pi_1(X,f(t))$ by $f_{\#}$ were still conjugate in $\pi_1(T,t)$, e.g. if $f_{\#}$ is onto, then you could make the same conclusion. 
Off the top of my head I don't have a counterexample for non-regular coverings but I doubt the same is true in general. The subgroups $H=p_{\#}(\pi_1(Y,y))$ and $H'=p_{\#}(\pi_1(Y,y'))$ will always be conjugate. But if $f_{\#}^{-1}(H)$ and $f_{\#}^{-1}(H')$ are not conjugate in $\pi_1(T,t)$, then they will be classified by different (likely non-homeomorphic coverings). To whip one up it should be enough to look for a group homomorphism $f_{\#}:G\to K$ which takes two non-conjugate subgroups in $G$ to two non-equal conjugate subgroups in $K$.
