Note that the fundamental group of the torus $T$ is given by $\pi_1(T)=\Bbb Z \times \Bbb Z =<a,b~|~aba^{-1}b^{-1}>$. By the Galois correspondence of covering spaces, I know that there is a unique (up to covering isomorphism) covering space of $T$ corresponding to (the conjugacy class of) the subgroup $<a^3, a^2b>$. Can I visualize this covering space using square diagrams? I want to describe the covering space explicitly, but I have no idea.
Visualizing a covering space of a torus
algebraic-topologycovering-spaces
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One subgroup of $\langle a, b\ |\ abab^{-1} = 1 \rangle $ that is isomorphic to $\mathbb{Z}\times \mathbb{Z}$ is the subgroup generated by $a$ and $b^2$. Indeed these elements commute: $ab^2 = ba^{-1}b = b^2 a$, and as this is a finite-index sub-group of a surface group it is also a surface group, so in particular it only has one relation. In general you could take $a^m$ and $b^{2n}$.
In an answer to a related question (Is there a non-trivial covering of the Klein bottle by the Klein bottle?) I gave a few families of subgroups and determined the total spaces of the corresponding cover space in those cases, maybe you will find it useful.
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.
Best Answer
Subdivide the universal covering space $\mathbb R^2$ in the usual manner as squares, with vertical lines $x=m$ and horizontal lines $y=n$ for integers $m,n \in \mathbb Z$.
To visualize the desired covering space, draw two vectors based at the origin: $v = \langle 3,0 \rangle$ corresponding to $a^3$; and $w = \langle 2,1 \rangle$ corresponding to $a^2 b$. Let $P$ be the parallelogram determined by the vectors $v,w$, which form two sides of $P$, the other two sides then being determined. Now glue opposite sides of $P$ to form a quotient space $S$. And as usual, gluing opposite sides of the unit square $Q = [0,1] \times [0,1]$ gives the covering space $T$.
You can then use the pattern of intersections of the parallelograph $P$ with unit squares $[m,m+1] \times [n,n+1]$ to define the desired covering map $Q \mapsto T$.
It is theoretically more straightforward to view this construction using orbit spaces of deck transformations. If I have $(a,b) \in \mathbb R^2$ let me use $\tau_{(a,b)}$ to represent the translation $\tau_{a,b}(x,y) (x+a,y+b)$. Thus we can think of $T$ as the quotient of $\mathbb R^2$ by the action of the deck group $\langle \tau_{(1,0)},\tau_{(0,1)} \rangle$ (with fundamental domain $[0,1] \times [0,1]$), so $$T = \mathbb R^2 / \langle \tau_{(1,0)},\tau_{(0,1)} \rangle $$ and we can think of $S$ as the quotient of $\mathbb R^2$ by the action of the deck group $\langle \tau_{(3,0)}, \tau_{(2,1)} \rangle$ (with fundamental domain $P$), so $$S = \mathbb R^2 / \langle \tau_{(3,0)}, \tau_{(2,1)} \rangle $$ This way, the desired quotient map $S \mapsto T$ can be precisely defined as the map $$S = \mathbb R^2 / \langle \tau_{(3,0)}, \tau_{(2,1)} \rangle \mapsto = \mathbb R^2 / \langle \tau_{(1,0)},\tau_{(0,1)} \rangle = T $$ that is induced by the identity map $\mathbb R^2 \mapsto \mathbb R^2$.