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If $f:[0,1]^2\mapsto\mathbb{R}$ is continuous, then $f$ is not injective

I start by ascerting that, since $[0,1]^2$ is compact, then there are $x_1,x_2\in[0,1]^2$ such that for every $x\in [0,1]^2$ $f(x_1)\leq f(x)\leq f(x_2)$; like this:

enter image description here

So, if I declare the sets $S$ and $S'$ like this:

enter image description here

The curves $f(S)$ and $f(S')$ will take all values between $f(x_1)$ and $f(x_2)$, since they will inherit continuity from $f$.

Thus, $f$ is not injective since $S$ and $S'$ both will have a point that maps to the same value between $f(x_1)$ and $f(x_2)$.

Is this proof correct? Thanks in advance.

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    $\begingroup$ Can you explain more what the two sets $S$ and $S'$ are? It seems very unclear... $\endgroup$
    – Anthony
    May 21, 2022 at 6:41
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    $\begingroup$ It is much easier to prove this using connectedness. $\endgroup$
    – Kavi Rama Murthy
    May 21, 2022 at 7:18
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    $\begingroup$ I think your argument is clear and correct, and gets to the point nicely. You can formalise it by defining the paths $S$ and $S’$ mathematically, but I rather think that that isn’t necessary in this simple setting $\endgroup$
    – Milten
    May 21, 2022 at 7:33
  • $\begingroup$ (And if you need to satisfy a teacher, you might wanna point to a specific pair of points that break injectivity) $\endgroup$
    – Milten
    May 21, 2022 at 7:35
  • $\begingroup$ I like your proof a lot. I think it is nice and clear. You might add that in the second step you choose some value $y$ strictly intermediate between $f(x_1)$ and $f(x_2)$, and that because of the intermediate value theorem both a point on $S'$ and on $S$ must get mapped to $y$. To be totally precise, you should also say that without loss of generality $f(x_1) < f(x_2)$. $\endgroup$
    – Nico
    May 21, 2022 at 9:14

2 Answers 2

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I'm writing this answer after taking the hint of @Kavi Rama Murthy.

Let $A:=[0,1]^2.$ Clearly, $A$ is connected. Suppose that $f$ is injective. The image of $f:A\to\Bbb{R}$, $f(A)$, is a closed interval, since $f$ is continuous. Let $x\in A$ be any point such that $f(x)$ is not one of the end-points of the image $f(A)$. Now, $A\setminus\{x\}$ is connected and the restriction of $f$ on this set, denoted by $f_{|_{A\setminus\{x\}}}$ is also continuous and injective. But the image of $f_{|_{A\setminus\{x\}}}$, $f(A)\setminus\{f(x)\}$ (because $f$ is injective), is not connected, leading to a contradiction.

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  • $\begingroup$ Why must injectivity imply the image is an interval? Continuity alone ensures this since $A$ is connected. $\endgroup$
    – FShrike
    May 21, 2022 at 8:46
  • $\begingroup$ I believe the statement $f(A\setminus\{x\})=f(A)\setminus f(x)$ is where you actually need injectivity $\endgroup$
    – FShrike
    May 21, 2022 at 8:48
  • $\begingroup$ @FShrike Ah, wrong placement of words. I'll make the required edits. Thank you. $\endgroup$
    – tmaj
    May 21, 2022 at 8:48
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Let $f:[0,1]^2\to\Bbb R$ an injective continuous function. Suppose an $a\in ([0,1]^2)^\circ$; then $f$ is still continuous in connected $[0,1]^2\setminus\{a\}$, but $f([0,1]^2\setminus\{a\})=f([0,1])\setminus \{f(a)\}$ is an interval without one point (by injectivity), so it is not connected, which is false, because the continuous image of a connected set is connected.

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