I'm wondering whether there is an invertible function $f: \mathbb{R} \to \mathbb{R}$ such that $f(-1)=0$, $f(0)=1$ and $f(1)=-1$. I think it's not but I'm missing a real proof.
The easiest would be to show such a function cannot be injective... But I don't see how? I don't see any other way of starting this either! Can you hint me please?
Thank you!
On
A continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ is invertible on an interval $I$ if, and only if, it is bijective, and it is bijective if, and only if, it is monotonous. Since your function $f$ is not monotonous on the interval [-1; 1] (it increases first, and then decreases), it cannot be invertible.
A graphical aid: If you draw a horizontal line, and the line intersects the graph of your function at more than just one point, then your function is not injective, and hence, not invertible.
However, you can shrink the interval, so as to make it injective in the smaller interval.
There is no invertible function like that which is continuous. Indeed, $f(0)=1$ and $f(1)=-1$ so by the Intermediate Value Theorem, there is $\xi \in (0,1)$ such that $f(\xi)=0 = f(-1)$.
There are discontinuous bijections of $\mathbb{R}$ which satisfy your conditions.