# What is wrong with this proof that isometries must be surjective?

Let $\phi : \mathbb{R}^2\rightarrow\mathbb{R}^2$ be an isometry. Suppose $\phi$ is not surjective, that is there exists some $v \in \mathbb{R}^2$ whose fiber $\phi^{-1}(v)$ is empty. Then by the pigeonhole principle there exist $u, u' \in \mathbb{R}^2$ where $u\neq u'$ which map to the same element $\phi(u)$. But then $\phi$ is not an isometry since $d(u,u') > d(\phi(u),\phi(u))=0$.

My issue is with using pigeonhole principle for uncountable sets, which feels flawed to me.

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Indeed, your problem is using the pigeonhole principle for infinite sets (not even uncountable). To wit, consider the map $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = 2x$. Since $f^{-1}(1)$ is empty, by your argument, $f$ must not be injective, which is clearly false.