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Show that the $3$-dimensional real projective space $\mathbb{R}P^3$ is homeomorphic to the lens space $L(2,1)$.

(I am not sure but the problem is probably from the book Knots and Links which is written by Rolfsen.)

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up vote 2 down vote accepted

In addition to explicitly identifying $\mathbb{RP}^3$ with the quotient of $\mathbb{S}^3$ by the antipodal map, here are is a second way of seeing that $L(2,1)\cong\mathbb{RP}^3$.

The lens space $L(p,q)$ can be constructed from a closed three-ball by gluing the southern hemisphere to the northern hemisphere by a $\frac{2\pi q}{p}$ twist. (This is illustrated quite well in Rolfsen.) In the case of $L(2,1)$, we glue south to north via a 180 degree twist. This is antipodal identification on the boundary sphere, which is another way of defining $\mathbb{RP}^3$.

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Definition. Think of $S^3$ as the unit sphere of the complex plane $\mathbb{C}^2$. Let $\tau: S^3 \rightarrow S^3$ be the homeomorphism given by $(z, w) \mapsto (ze^{2\pi i/ p}, we^{2\pi i q/p})$. Then since $\tau$ is periodic of $p$, it creates a $\mathbb{Z}/p$ group action on $S^3$. We identify the points $x, y$ of $S^3$ if $x=\tau^k(y)$ for some $k$. The resulting space is the lens space $L(p, q)$.

$\mathbb{R}P^3$ can be seen as a quotient space of $S^3$ with antipodal points identified. In the above definiton, when $p=2$ and $q=1$, we only identify $(z, w)$ with $(-z, -w)$ that are antipodal points. Hence we are done.

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