Why the lens space L(2,1) is homeomorphic to $\mathbb{R}P^3$? According to one definition of lens space $L(p,q)$, which is gluing two solid tori with a map $h:T^2_1 \rightarrow T^2_2$. And $h(m_1)=pl_2+qm_2$, $l_i$ means longitude and $m_i$ means meridian of the boundary torus. I cannot understand why $L(2,1)$ is homeomorphic to $\mathbb{R}P^3$. Can someone give me some hints? Thank you.

The curve $\gamma= m_2 + 2 \ l_2$ bounds a Moibus strip $M$ inside the the solid torus $T_2\subset L(2,1)$ (can you see it?) and a two-disk $D$ inside $T_1 \subset L(2,1)$. The union  $S= D \cup M$ is a closed non-orientable surface  homeomorphic to $\mathbb{RP}^2$, and you can easily see that its complement in $L(2,1)$ is nothing but a 3-ball. 
Since $L(2,1)$ is orientable while $\mathbb{RP}^2$ is not, the normal bundle of $S$ inside $L(2,1)$ is a non-trivial line bundle. Hence, a normal neighborhood of $S \subset L(2,1)$ is homeomorphic to $\mathbb{RP}^2 \tilde{\times} I$ (the interval bundle associated to the unique non-orientable line bundle over $\mathbb{RP}^2$), and we can conclude that $L(2,1)$ is obtained from $\mathbb{RP}^2 \tilde{\times} I$ by filling its boundary 2-sphere with a 3-ball. 
On the other hand, you can see directly that the normal neighborhood of $\mathbb{RP}^2 \subset \mathbb{RP}^3$ is homeomorphic to $\mathbb{RP}^2 \tilde{\times} I$ and that its complement is homeomorphic to a 3-ball. Namely, think $\mathbb{RP}^3$ as the unit ball $B^3(0,1)/\sim$ with the antipodal relation on its boundary, and decompose it as the union of the ball $B^3(0,1/2) \subset B^3(0,1)/\sim$ and its complement (homeomorphic to a non-trivial interval bundle over $\mathbb{RP}^2 = \partial B^3(0,1)/\sim$).
Hence, both $L(2,1)$ and $\mathbb{RP}^3$ are obtained from $\mathbb{RP}^2 \tilde{\times} I$ by filling its boundary 2-sphere with a 3-ball. Now, in order to conclude, we only need an important basic fact from differential  topology: a 2-sphere boundary component of 3-manifold can be capped  uniquely with a 3-ball.
