Show that there is no covering map \[ p:T^2=S^1\times S^1\longrightarrow\RP^2. \]
Solution
1. A covering map \(p:T^2\to\RP^2\) would induce an injective homomorphism\[ p_*:\pi_1(T^2)\hookrightarrow\pi_1(\RP^2). \]
A covering-space projection induces an injection on fundamental groups of path-connected based spaces.
2. The groups in 1 are\[ \pi_1(T^2)\cong\ZZ^2, \qquad \pi_1(\RP^2)\cong\ZZ/2\ZZ. \]
The first follows from \[ T^2=S^1\times S^1 \] and the product formula for fundamental groups. The second is the standard computation from the universal double cover \[ S^2\to\RP^2. \]
3. No injective homomorphism \(\ZZ^2\to\ZZ/2\ZZ\) exists.
The image of any such homomorphism is a subgroup of the two-element group \(\ZZ/2\ZZ\), hence has at most two elements. The group \(\ZZ^2\) is infinite, so an injective map is impossible.
4. No covering map \(T^2\to\RP^2\) exists.
Otherwise 1 and 2 would produce the injection ruled out by 3.