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I would like to prove that the following map is injective

$$ f: \mathbb{S}^1 \to \mathbb{RP}^1 = \mathbb{R}^2 / \sim $$

$$ f(x) = \begin{cases} (1,0)&\text{ if } x \text{ is south } \\ [x]&\text{otherwise} \end{cases} $$
where for $x,y \in \mathbb{R}^2 \setminus \lbrace (0,0) \rbrace$ we define $x \sim y$ by $x=\lambda y$ for some $\lambda \neq 0$.

Here is what I have done so far:

Let $f(x)=f(y)$ for some $x, y \in \mathbb{S}^1$, that means $ [x] = [y] $, since these classes are equal, this happens if and only if $x \sim y$, so $x = \lambda y$ for some non zero scalar $\lambda$ but due to symmetry $y \sim x$, that is $y = \lambda x$, then $x = y$ when $\lambda=1$. Is this resasoning correct?

Anne Bauval
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Daniel R.S
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  • No, and this function isn't injective. If $x \sim y$, $x = \lambda y$ (but note that your error here is that $y$ doesn't need to be $\lambda x$--by symmetry it's indeed equal to a scalar times $x$ but that scalar is not necessarily the original $\lambda$). Since $x$ has length 1, you obtain, by taking length of both sides, $1 = |\lambda|$. So $\lambda = \pm 1$. Hence there are two points on $S^1$ that yield the same value under $f$, $x$ and $-x$. – kamills May 13 '24 at 18:58
  • Oh thanks, and what if I consider $\mathbb{S}^1$ as $\lbrace x^2 + (y-1)^2 \rbrace=1$ maybe I could get rid of the problem of antipodal points? Just wondering :b – Daniel R.S May 13 '24 at 19:06
  • This (if completed by some definition of $f$) would be another question hence another post. For a more standard map, see https://math.stackexchange.com/questions/311196 Now that your error was explained, can your post be considered as a duplicate of the latter? – Anne Bauval May 13 '24 at 19:14
  • Yes, thanks for the clarification – Daniel R.S May 13 '24 at 20:00

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