I'm looking for a counterexample. I already proved that if $(X,d_X)$ and $(Y,d_Y)$ are metric spaces and $X$ is complete and $f:X\to Y$ is a bijection that $Y$ is complete if $f^{-1}$ is uniform continuous. I know that if $f^{-1}$ isn't uniform continuous $Y$ isn't complete. I just can't find a counterexample.
I tried to look for $f:\mathbb{R}\to\mathbb{Q}$ or something but it's never right. I found online that $f:\mathbb{R}\to(0,1):x\mapsto1/(1+2^{-x})$ is a counterexample but i don't really see how. Can someone maybe give an easier example or explain me why this is a great counterexample?
Edit: i'm looking for a function with f is uniform continuous and f^-1 is normal continuous, where X is complete but Y isn't
$f(x)=\frac x {1+|x|}$ is a homeomorphism from $\mathbb R$ onto $(-1,1)$. The domain is complete but the range is not.
Alternatively use the homeomorphism $\arctan x$ from $\mathbb R$ onto $(-\pi /2 ,\pi /2)$.
These two functions are uniformly continuous bijections.