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I have to prove that if $f: [0,1] \to [0,1]$ is a bijection and continuous, only two things happen that $f(0)=0$ and $f(1)=1$ or $f(1)=0$ and $f(0)=1$.

This was my idea, let $f(0)=a$ and $f(1)=b$, then exists $a',b' \in [0,1]$ such that $f(a')=a$ and $f(b')=b$, but I don't know how to do

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  • $\begingroup$ @Fawkes4494d3 I forgot to write that f is continuous $\endgroup$ – Alejandra Benítez Sep 17 at 3:12
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This is false. Define $$ f(x) = \begin{cases} 0.25 & x = 0 \\ 0.75 & x = 1 \\ 0 & x = 0.25 \\ 1 & x = 0.75 \\ x & \text{otherwise} \end{cases} $$ That's a bijection but doesn't have the claimed property.

If you ask about continuous bijections, then the claim is true, however.

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  • $\begingroup$ I forgot to write that f is continuous $\endgroup$ – Alejandra Benítez Sep 17 at 3:12
  • $\begingroup$ This answer (and also the deleted answer) completely answer your question as it originally posted. I don't understand why somebody down voted for them. $\endgroup$ – Bumblebee Sep 17 at 3:26
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    $\begingroup$ Indeed. Once you have multiple correct answers to a mistaken question, it's probably nicer to re-ask as a NEW question, so that the correct answers don't end up looking stupid. After all, the folks who answered those are the ones who did something right. $\endgroup$ – John Hughes Sep 17 at 3:29
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Suppose $f(0) = v$ with $v \neq 0$ and $v \neq 1$, that is $v \in (0, 1)$. Since the function is surjective, there must be some $a, b \in (0, 1]$ with $f(a) = 0$ and $f(b) = 1$. By the intermediate value theorem, there is $c \in (a, b)$ (or $c \in (b, a)$ if $b < a$) with $f(c) = v$ and the function is not injective, a contradiction. A similar argument also applies for $f(1)$.

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The continuous image of a compact set is compact; in the reals, Heine-Borel says that means closed and bounded. Hence such a function achieves its maximum. So there's a point $c$ with $$ f(c) = 1. $$ If $0 < c < 1$, then pick $a = c/2$; $b = \frac{1 + c}{2}$. Then evidently $ 0 < a < c < b < 1$, and $f(a) < 1$ and $f(b) < 1$.

Now the continuous image of a connected set is connected, so $f([a, c])$ is a connected interval $[A, 1]$, where $A < 1$; similarly, $f([c, b])$ is a connected interval $[B, 1]$. Letting $u = \max{A,B} < 1$, we have $u \in f([a, c))$ and $ u \in f( (c, b])$, hence $f$ is not injective.

Thus the assumption that $f(1)$ is strictly between $0$ and $1$ is false, so $f(1) = 0$ or $f(1) = 1$. A similar argument shows the same thing for $f(0)$.

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  • $\begingroup$ $A$ and $B$ are not necessarily less than 1 as the function can be constant near the maximum (though the rest of the argument stands anyway) $\endgroup$ – MBW Sep 17 at 3:40
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    $\begingroup$ The function cannot be constant near the max, because it's a bijection --- the max may be assumed at at most one argument. But I should have noted that. $\endgroup$ – John Hughes Sep 17 at 10:57
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In fact, there's a fact about $\underline{\text{continuous functions which are bijections}}$ the idea of the proof of which comes in handy:

If $f:\Bbb{R}\to\Bbb{R}$ is bijective and continuous, $f$ is strictly monotone.

Proof: Suppose $f$ is not strictly monotone, that is, without loss of generality, $\exists a,b,c\in\Bbb{R}, a<b<c$ such that $f(a)<f(b),f(b)>f(c)$,
call $A=f(a),B=f(b), C=f(c)$, i.e. $A,C<B$
by the Intermediate value theorem, for a $D$ such that $D\in(C,B)$ and $D\in (A,B)$,
$\exists d_1 \in (b,c),$ such that $f(d_1)=D$
$\exists d_2 \in (a,b),$ such that $f(d_2)=D$
but $d_2 \in (a,b), d_1\in (b,c)\implies d_1\ne d_2$ but $f(d_1)=f(d_2)\implies$ $f$ is not a bijection, which is a contradiction. So $f$ is strictly monotone.
($\because $ $f$ is bijective, no equality signs i.e [ $\le,\ge$ ] among $f(a),f(b),f(c)$ have been considered after assuming the negation of $f$ being strictly monotone.)


This fact can be directly used to prove your claim.

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    $\begingroup$ For which fact there's an easier proof: math.stackexchange.com/questions/170147/…. $\endgroup$ – Michael Hoppe Sep 17 at 11:44
  • $\begingroup$ @MichaelHoppe ah yes :D that is easier, and it's a more helpful proof even when the problem does not offer much geometric/graphical intuition. $\endgroup$ – Fawkes4494d3 Sep 17 at 12:00

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