Mapping cone z to z^2 on S1 is RP2

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I want to show that the mapping cone of $z \mapsto z^2$ on $S^{1}$ is homeomorphic to $\mathbb{RP}^{2}$.

My thought was:

1) $\mathbb{RP}^{2}$ is indeed $S^{2}$ with identifying each pair of antipodal points as one point.

2) We can assign "cross sections" of positive latitude of cone to each corresponding "cross sections" with same latitude of the upper hemisphere of $S^{2}$.

3) In the case of section of zero latitude, for any $x \in S^{1}$, there are two distinct (indeed, antipodal) points $y$, $-y$ in $S^{1}$ such that $y^{2}=(-y)^{2}=x$, so we are done.

I guess that this approach is right, but I don't know how to refine them to a rigorous proof. Please help me. Thanks.