# Find all connected covering space of $\mathbb RP^2\vee \mathbb RP^2$

This is exercise 1.3.14 in page 80 of Hatcher's book Algebraic topology.

To move this question out of the unanswered list, I put my solution in answer.

• The list of cyclic subgroups of order 2 seems incomplete: $bab$ will generate one, in the same way that $aba$ will. Dec 12, 2019 at 16:02
• Thanks for pointing this out. If we don't care about the choice of basepoint, there're only two covering spaces corresponding to subgroups isomorphic to cyclic group of order $2$. I'll modify the whole proof. Dec 13, 2019 at 4:29

In the following pictures, green dots means basepoints, black curve means its two endpoints are attached. Covering map maps blue $$S^2$$ to $$X_1$$ and red $$S^2$$ to $$X_2$$.

Let $$X_1$$ and $$X_2$$ denote the first and second copy of $$\mathbb RP^2$$.

$$\pi_1(X_1)=\mathbb Z_2=\langle a \rangle,\ \pi_1(X_2)=\mathbb Z_2=\langle b \rangle$$.

$$1$$. For trivial subgroup $$1$$, it corresponds to the the universal cover, i.e. the infinite chain of $$S^2$$.

$$2$$. For subgroup isomorphic to infinite cyclic group $$\mathbb Z$$, it is generated by $$(ab)^n$$ or $$(ba)^n$$ of index $$2n$$ $$(n \geqslant 1)$$ and it corresponds to a "necklace" of $$2n$$ copies of $$S^2$$.

$$3$$. For subgroup isomorphic to $$\mathbb Z_2$$, it's generated by $$(ab)^{m}\cdot a$$ or $$(ba)^{m}\cdot b$$ $$(k\geqslant 0)$$ and it corresponds to $$\mathbb RP^2$$ attached to an infinite chain of $$S^2$$.

$$4$$. For subgroup isomorphic to the infinite dihedral group $$\mathbb Z_2 * \mathbb Z_2$$, it's generated by $$(ab)^n$$ and $$(ab)^m \cdot a$$ $$(m\leqslant n)$$ and it corresponds to a finite chain of $$S^2$$'s with both ends attached an $$\mathbb RP^2$$.

• Beautiful pictures, how did you make them? Feb 9, 2020 at 10:02
• @AlessandroCodenotti Thank you. Actually I draw them in OneNote by hand, and it took a lot of time, but I think that's worthy if it can really benefit others. The answer to this question is not hard, but most of the answers are descriptive. Feb 9, 2020 at 10:42
• I understand given a group $G$, one can always construct a universal cover $\tilde{X}_G$. But is the covering space you draw corresponding to each subgroup of $G$ universal? It seems, for example, the second covering space corresponding to the subgroup $\langle (ab)^n \rangle$ is not simply connected. Feb 14 at 3:37
• @ZelongLi Why should it be simply connected? Only the covering space corresponding to the trivial subgroup should be simply connected. May 26 at 3:53