Fundamental Group of a Quotient of an Annulus

(For reference, this is problem 10-15 in Lee's Introduction to Topological Manifolds, 2ed)

We are given the annulus $$Q=\{z\in\mathbb{C}|1\le|z|\le 3\}$$ and the quotient corresponding to the relation $$z\sim -z,\quad\forall z\in\partial Q.$$ Let $\widetilde{Q}=Q/\sim$.

My intuition tells me that $\widetilde Q\cong \mathbb{R}\mathrm{P}^2\#\mathbb{R}\mathrm{P}^2$, which would give us the corresponding fundamental group $\langle a,b|a^2b^2=1\rangle$ using a result on the correspondence between fundamental groups and polygonal presentations for compact surfaces.

Alternatively, using $U=B_{3}(0)\cap\widetilde{Q}$ and $V=\widetilde{Q}\setminus S^1$ as my two open sets, I can also apply Seifert-Van Kampen where $U$ and $V$ both admit strong deformation retractions onto $\mathbb{R}\mathrm{P}^2$ and $\pi_1(U\cap V)=\mathbb{Z}$.

Synthesizing everything and glossing over some of the details, it seems that this route leads me to the conclusion that $$\pi_1(\widetilde{Q})\cong\langle a,b|a^2=b^2=ab=1\rangle$$ which is certainly not the same group as before (this one is very finite, for instance).

Both of these methods should work, so I must have some faulty reasoning somewhere. Can someone please help me locate where I'm going wrong?

• In the alternative approach, the open sets $U$ and $V$ retract onto $\mathbf{RP}^{1}$ (the central circle of a Moebius strip), not $\mathbf{RP}^{2}$. Nov 27, 2016 at 3:02
• Sorry, are you talking about $\widetilde{Q}$? I don't believe it is. It is certainly an annulus before the identifications are made.