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.

Here's a sort of diagrammatic argument from an old homework assignment:
Construct $\mathbb{R}P^3$ as the quotient of $B^3 \subset \mathbb{R}^3$ under the antipodal map $a: \partial B^3 \to \partial B^3$. Let $K$ be the knot in $\mathbb{R}P^3$ obtained as the quotient of the vertical segment $V=\{(0,0,z) \in \mathbb{R}^3 : -1 \leq z \leq 1\}$ in $B^3$. As depicted in the figure below, a normal neighborhood $N(K)$ of $K \subset \mathbb{R}P^3$ is a solid torus.  We claim its complement is also a solid torus. As depicted, there is a simple closed curve in $\partial N(K)$ that bounds an embedded disk $(D,\partial D) \subset (\mathbb{R}P^3 \setminus \mathring{N}(K), \partial N(K))$. Manipulating the identification diagram (not shown, but achieved by cutting $B^3 \setminus V$  along the disk and gluing via the antipodal map on $S^2 \setminus \{(0,0,\pm 1)\}$), we can see that the complement of this disk in $\mathbb{R}P^3 \setminus N(K)$ is a 3-ball. It follows that $\mathbb{R}P^3 \setminus \mathring{N}(K)$ is a solid torus.  As discussed below the figure, we can apply a half-twist to $\partial N(K)$ to identify it with a standard torus, taking $\partial D$ to a $(2,1)$-curve in $\partial(S^1 \times D^2)$. This decomposition corresponds to a homeomorphism from $\mathbb{R}P^3$ to the lens space $L(2,1)$. $\square$

Left: $\mathbb{R}P^3$ is depicted as a quotient of the closed 3-ball $B^3$ via the antipodal map $a: \partial B^3 \to \partial B^3$. The vertical line depicts the knot $K \subset \mathbb{R}P^3$. Middle: The knot's normal neighborhood $N(K)$ situated inside of $\mathbb{R}P^3$. The boundary of $N(K)$, seen here as a dotted cylinder, is a torus. We can visualize a deformation retract of $\mathbb{R}P^3 \setminus N(K)$ to the equatorial $\mathbb{R}P^1=S^1$ by taking successively "larger" neighborhoods $N(K)$. Right: We see a simple closed curve in $\partial N(K)$ which bounds a disk in $\mathbb{R}P^3 \setminus K$. The antipodal map sends the "upper circle" to the "lower circle" with a rotation by $\pi$, so the depicted curve becomes a $(2,1)$-curve in $\partial N(K)$.
