no bijective continuous map from $(-1,1) \to (-1,1]$

Question

How to show there is no bijective continuous map from $(-1,1) \to (-1,1]$ ?

I'm clueless, any help would be appreciated...

Answer 1

Here's a proof not relying on connectedness.

Suppose $f:(-1,1)\to (-1,1]$ is continuous and bijective.

There exists $x\in (-1,1)$ such that $f(x)=1=\max_{t\in (-1,1)}f(t)$. But since $f(a)<1$ for all $a<x$ and $f(b)<1$ for all $b>x$, it follows from continuity that there exist $a<x<b$ such that $f(a)=f(b)$, contradiction.

Answer 2

Hint: Since the map is a bijection, some number has to map to 1. Which number? What happens to numbers greater than and less than that number?

Answer 3

Let $x_0\in(-1,1)$ with $f(x_0)=1$ ($f$ is surjective). Consider $A=(-1,x_0)$ and $B=(x_0,1)$; then $$ \lim_{x\to x_0^-}f(x)=1 $$ and, since the image of $A=(-1,x_0)$ under $f$ is an interval, we have $f(A)=(y_0,1)$ or $f(A)=[y_0,1)$, because $1$ is a limit point of $f(A)$ and doesn't belong to $f(A)$ by injectivity.

Similarly, from $$ \lim_{x\to x_0^+}f(x)=1 $$ we obtain that, if $B=(x_0,1)$, then $f(B)=(z_0,1)$ or $f(B)=[z_0,1)$.

Note that $y_0<1$ and $z_0<1$ and that $(y_0,1)\cap(z_0,1)\ne\emptyset$. This easily provides the contradiction you were looking for.