Let $G = \pi_1(X,x_0)$ be the fundamental group of a topological space $X$.
Let $H_1,H_2$ be subgroups of $G$, and by the correspondence between between subgroups of $G$ and covering spaces of $X$, we have two covering spaces:
$p_1:(X_1,x_1)\to (X,x_0)$ such that $H_1 = (p_1)_*(\pi_1(X_1,x_1))$
$p_2: (X_2,x_2)\to (X,x_0)$ such that $H_2 = (p_2)_*(\pi_1(X_2,x_2))$
I'm interested in the covering space that correspond to the subgroup $H_1\cap H_2$. I think it is the space
$$Y = \{(a_1,a_2)\in X_1\times X_2|p_1(a_1) = p_2(a_2)\}$$
but I'm not sure how to prove it.
One of the ideas I had was to look at the "Galois correspondence" between subgroups of $G$ and the covering spaces of $X$. categorically speaking this correspondence is in the form of adjoint functors between the order categories of covering spaces and of subgroups of $G$ (with reverse ordering) . Thinking like that, $H_1\cap H_2$ is the colimit of the discreet diagram $H_1, H_2$ in the latter category. So we are looking for the colimit of the discreet diagram $X_1, X_2$ in the category of covering spaces of $X$. That is where I fail to continue.
I will be glad for any form of help.
Best Answer
Introductory rambling: the space $Y$ together with the obvious projections $\pi_1\colon Y\rightarrow X_1$ and $\pi_2\colon Y\rightarrow X_2$ is the fiber product of the maps $p_1\colon X_1\rightarrow X$ and $p_2\colon X_2\rightarrow X$. Categorically speaking, these maps constitute a pullback diagram in the category $\mathbf{Top}$. It is well-known that this corresponds to the product of $p_1\colon X_1\rightarrow X$ and $p_2\colon X_2\rightarrow X$ in the slice category $\mathbf{Top}/X$. Now, the category $\mathbf{Cov}(X)$ of covering spaces over $X$ is a full subcategory of $\mathbf{Top}/X$. The fact that $p\colon Y\rightarrow X$ is itself a covering space (I assume you have already confirmed this) implies that it is also the product of $p_1\colon X_1\rightarrow X$ and $p_2\colon X_2\rightarrow X$ in $\mathbf{Cov}(X)$ (the inclusion of a full subcategory reflects limits). This is nice and all, but not quite what we want as unpointed covering spaces only correspond to conjugacy classes of subgroups on the other end.
Thus, let us use your correct observation that the category $\mathbf{Cov}(X,x_0)$ of pointed covering spaces is eqiuvalent to the posetal category of subgroups of $\pi_1(X,x_0)$. In this context, the covering corresponding to $H_1\cap H_2$ is a product of $p_1\colon(X_1,x_1)\rightarrow(X,x_0)$ and $p_2\colon(X_2,x_2)\rightarrow(X,x_0)$ in the category $\mathbf{Cov}(X,x_0)$. It suffices to check that $p:(Y,y_0)\rightarrow(X,x_0)$ together with the projection maps $\pi_1\colon(Y,y_0)\rightarrow(X_1,x_1)$ and $\pi_2\colon(Y,y_0)\rightarrow(X_2,x_2)$ satisfies the universal property. To this end, let $q\colon(Z,z_0)\rightarrow(X,x_0)$ be a covering space and $\rho_1\colon(Z,z_0)\rightarrow(X_1,x_1)$ and $\rho_2\colon(Z,z_0)\rightarrow(X_2,x_2)$ be morphisms of covering spaces, i.e. $p_1\circ\rho_1=q=p_2\circ\rho_2$. Then, by the universal property of the pullback, there exists a continuous map $\tilde{q}\colon Z\rightarrow Y$, such that $\pi_1\circ\tilde{q}=\rho_1$ and $\pi_2\circ\tilde{q}=\rho_2$. This forces $\tilde{q}(z_0)=y_0$ (either by the explicit construction or by another application of the universal property), whence $\tilde{q}\colon(Z,z_0)\rightarrow(Y,y_0)$ is a morphism of pointed coverings. It is necessarily unique, since the category is posetal. The claim follows. (The point is that compatibility with basepoints is automatic in this scenario. Similarly, we obtain that the forgetful functor $\mathbf{Cov}(X,x_0)\rightarrow\mathbf{Cov}(X)$ creates products.)
Alternatively, let me offer a concrete argument to balance the abstraction. An element $[\gamma]\in\pi_1(X,x_0)$ is contained in $p_{\ast}(\pi_1(Y,y_0))$ if and only if there exists a closed lift $\tilde{\gamma}\colon I\rightarrow Y$ based at $y_0$ such that $p\circ\tilde{\gamma}=\gamma$. If such a lift exists, then $\pi_1\circ\tilde{\gamma}$ is a closed lift of $\gamma$ through $p_1$ based at $x_1$ and $\pi_2\circ\tilde{\gamma}$ is a closed lift of $\gamma$ through $p_2$ based at $x_2$, forcing $[\gamma]\in(p_1)_{\ast}(\pi_1(X_1,x_1))\cap(p_2)_{\ast}(\pi_2(X_2,x_2))=H_1\cap H_2$. Conversely, if $[\gamma]\in H_1\cap H_2=(p_1)_{\ast}(\pi_1(X_1,x_1))\cap(p_2)_{\ast}(\pi_2(X_2,x_2))$, there exist closed lifts $\tilde{\gamma}_1$ of $\gamma$ through $p_1$ based at $x_1$ and $\tilde{\gamma}_2$ of $\gamma$ through $p_2$ based at $x_2$. Then, $t\mapsto(\tilde{\gamma}_1(t),\tilde{\gamma}_2(t))$ is a closed lift of $\gamma$ through $p$ based at $y_0$ (alternatively, this is again the universal property of the pullback). In total, $p_{\ast}(\pi_1(Y,y_0))=H_1\cap H_2$, as desired.
It might be instructive to note that these two proofs are essentially saying the same things, just in different languages.