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In an oral exam I passed yesterday evening I was asked a question similar to the following: Does there exist an application $f:\mathbb{R}^2\to\mathbb{R}$ that is continuous and injective? I'd like to know how to render my solution perfectly rigorous as to why it is impossible.

The idea is to say since $f$ is continuous and $\mathbb{R}^2$ is simply connected, it follows that $f(\mathbb{R}^2)=(a,b)$ where $-\infty\leq a<b\leq+\infty$. by taking $\Omega=B_f((0,0),r)$ the closed ball of radius $r>0$ and centre $(0,0)$, $f(\Omega)=I\subset(a,b)$ where $I$ is an interval by continuity. The idea is now to consider two points $(A,B)\in(\mathbb{R}^2\backslash\Omega)^2$ and observe that there are two cases:

  • Either $[A,B]\cap\Omega$ is empty, in which cases by continuity on the segment $[A,B]$, we have that $f(A)$ and $f(B)$ map to the same side of $I$ and cannot be inside of it by injectivity.
  • Either the intersection is non-empty, in which case once again by continuity of $f$ on the segment $[A,B]$ we have that $I\subset[f(A),f(B)]$, ie $f(A)$ and $f(B)$ do not map to the same side of $I$.

It is this second case that interests us as, if we take a point $C\in\mathbb{R^2}$ far enough, well chosen such that the intersections $[A,C]\cap\Omega$ and $[C,B]\cap\Omega$ are empty, we have a contradiction, as by continuity on the one hand $A$ and $C$ must be mapped to the same side of $I$ and yet $C$ and $B$ must be mapped to the same side f $I$, which is impossible since $A$ and $B$ aren't mapped on the same side of $I$.

This proof is incredibly visual, and I would like some help in formalising the proof (which may actually be wrong! This solution is one I adapted from another problem that i actually got during my exam, which was validated by the examiner).

The question I was asked was Does there exist an application $f:\mathbb{R}^2\to\mathbb{R}$ that is simultaneously surjective, continuous such that $f^{-1}(\{0\})$ is compact?

J.J.T
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    An easy solution is that the restriction of the map to the circle is a homeomorphism onto its image due to compactness and Hausdorfness. But this is impossible since any path connected sub space of the real numbers is necessarily simply connected. For the question you were posed, consider the distance function from the origin – Exit path Oct 01 '24 at 20:33
  • @Exitpath Even simpler, without referring to 1-connectivity: Since $f$ is injective, $f(\mathbb{R}^2 \setminus {p}) = f(\mathbb{R}^2) \setminus {f(p)}$ for any $p \in \mathbb{R}^2$. But for a suitable choice of $p$, the right-hand side is disconnected, contradiction. In a sense, OPs proof idea resolves to this. – Ben Steffan Oct 01 '24 at 20:37
  • @BenSteffan Your comment should be an answer. – Anne Bauval Oct 01 '24 at 20:38
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    This is an unnecessarily complicated solution. The question itself was asked several times before on MSE. – Moishe Kohan Oct 01 '24 at 20:38
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    @AnneBauval My comment does not answer the question posed, which is asking for help with formalizing the given proof idea. – Ben Steffan Oct 01 '24 at 20:39
  • @Jacques See https://math.stackexchange.com/a/568919 and https://math.stackexchange.com/questions/116350 – Anne Bauval Oct 01 '24 at 20:45

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