Let \(T = S^1 \times S^1\), the torus. There is an isomorphism of \(\pi_1(T, b_0 \times b_0)\) with \(\mathbb{Z} \times \mathbb{Z}\) induced by projections of \(T\) onto its two factors.
Find a covering space of \(T\) corresponding to the subgroup of \(\mathbb{Z} \times \mathbb{Z}\) generated by the element \(m \times 0\), where \(m\) is a positive integer.
Find a covering space of \(T\) corresponding to the trivial subgroup of \(\mathbb{Z} \times \mathbb{Z}\).
Find a covering space of \(T\) corresponding to the subgroup of \(\mathbb{Z} \times \mathbb{Z}\) generated by \(m \times 0\) and \(0 \times n\), where \(m\) and \(n\) are positive integers.
Solution
Write \[ q:\mathbb R\to S^1,\qquad q(t)=e^{2\pi i t}. \] Under the standard identification \(\pi_1(T)\cong\mathbb Z\oplus\mathbb Z\):
Define \[ p_m:S^1\times\mathbb R\to S^1\times S^1,\qquad p_m(z,t)=(z^m,q(t)). \] This is a covering map. Since \(\pi_1(S^1\times\mathbb R)\cong\mathbb Z\), the generator represented by the first circle maps to \((m,0)\). Thus the corresponding subgroup is \[ \langle(m,0)\rangle. \]
The universal covering \[ p:\mathbb R^2\to T,\qquad p(s,t)=(q(s),q(t)) \] has simply connected total space, so its associated subgroup is trivial.
Define \[ p_{m,n}:T\to T,\qquad p_{m,n}(z,w)=(z^m,w^n). \] Then \[ (p_{m,n})_*:\mathbb Z^2\to\mathbb Z^2,\qquad (r,s)\mapsto(mr,ns), \] so its image is precisely \[ \langle(m,0),(0,n)\rangle. \] Hence these three maps realize the required covering spaces.