Question on the injectivity of a function. Consider the map $f\colon\mathbb{S}^1\to\mathbb{S}^1$ defined as $f(z)=z^2$, where $\mathbb{S}^1$ is the unit circle, $$\mathbb{S}^1=\{z\in\mathbb{C}:|z|=1\}.$$ On $\mathbb{S}^1$ we define the following equivalence relation $$x\sim y\quad\iff\quad x=y\;\text{or}\;x=-y.$$ Consider the map $\overline{f}\colon\mathbb{S}^n/\sim\to\mathbb{S}^1$ defined as $\overline{f}\big([z]\big)=f(z)$, for $z\in\mathbb{S}^1.$

Question. Is the map $\overline{f}$ injective?



My attempt. Yes. Indeed given $z_1,z_2\in\mathbb{S}^1$, if $\overline{f}\big([z_1]\big)=\overline{f}\big([z_2]\big)$, then $f(z_1)=f(z_2)$, so $z_1^2=z_2^2,$ therefore $z_1=\pm z_2$ and then $z_1\sim z_2$, then $[z_1]=[z_2].$ It's correct?

Thanks!

Yes, you're correct. In fact, when we're showing $\overline{f}$ is well defined its injectivity naturally appears: for $z_1,z_2\in\mathbb{S}^1$
\begin{align*}
[z_1]=[z_2] & \iff z_1=\pm z_2 \\
& \iff z_1^2 =z_2^2 \\
& \iff f(z_1)=f(z_2) \\
& \iff \overline{f}([z_1])=\overline{f}([z_2])
\end{align*}
