# Universal cover of a connected sum

Maybe it's an easy question: How can we find the universal cover of the connected sum of tori or projective plans? Is there a general method?

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See this answer by Ryan Budney. – Juan S Aug 29 '12 at 8:46
For the surfaces of genus $g > 0$ the universal cover is the hyperbolic plane, this is a classical result. I know that this is not really an answer to your question so I stated it as a comment. – mland Aug 29 '12 at 10:00
Have you a reference? – Seirios Aug 29 '12 at 15:36

Let $X_1$ (resp. $X_2$) be a sphere and $D_1$, $D_1'$ (resp. $D_2$, $D_2'$) two diametrically opposed disks on $X_1$ (resp. $X_2$). Then, set $\tilde{X}_1= X_1 \backslash (D_1 \cup D_1')$ and $\tilde{X}_2=X_2 \backslash (D_2 \cup D_2')$. Finally, define $\tilde{X}$ by $(\tilde{X}_1 \coprod \tilde{X}_2 )/ \sim$, where $\sim$ identifies $D_1$ with $D_2$, and $D_1'$ with $D_2'$.
Then, $X_i$ is the universal cover of a projective plan (and the projection is given by identifying two diametrically opposed points on $X_i$) so $\tilde{X}$ is a $2$-sheet covering space of the connected sum of two projective plans. But it is easy to see that $\tilde{X}$ is homeomorphic to $\mathbb{T}$, so the universal cover of the sum of two projective plans is homeomorphic to $\mathbb{R}^2$. (In fact, the result doesn't depend on the number of projective plans we add.)
However, the projection from $\mathbb{R}^2$ onto the connected sum is difficult to explicit because the homeomorphism between $\tilde{X}$ and the torus is itself difficult to explicit. But this method shows that the fundamental group of the connected sum of two projective plans contains a subgroup of index $2$ isomorphic to $\mathbb{Z}^2$.