$f: [0,1] \to [0,1]$ is a continuous bijection $\implies\{f(0),f(1)\}=\{0,1\}$ 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
 A: 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.
A: 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 $f(1)$.
A: 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 $f(0)$.
A: 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.
