Let $(X,d_x)$ be a complete metric space and $(Y,d_y)$ just a metric space. We know that $f:X \rightarrow Y$ is a bijection.
Case 1/ $f$ is continuous and $f^{-1}$ is uniformly continuous, is $Y$ complete:
Take a random cauchy sequence $y_n$ in $Y$. Then $f^{-1}(y_n)=x_n$. Since $X$ is complete and because the inverse is uniformly continuous, we know that $x_n \rightarrow x$ and thus also $f(x_n) \rightarrow f(x)$ and therefore $y_n \rightarrow f(x)=y$. Thus $Y$ complete.
Case 2 $f$ is uniformly continuous and $f^{-1}$ is continuous, is $Y$ complete:
I was searching for a counter example. I thought the function $f(x)=\sqrt{x}$. When you take the cauchy row $y_n=\sqrt{n}$ the row $f^{-1}(y_n)=n$ which is not a cauchy row and therefore we can not conclude that $y_n$ must be complete. But I can not find a fitting space so that my function stays a bijection.
Can anybody help me to check my case 1 and give help with the second case? I think my first case might be wrong because I'm not sure is $f(x) \in Y$ is always correct but I can not find a counter example.
Thanks in advance.
Best Answer
Your answer for 1) is correct (but writing needs some improvements).
For 2) consider $f: \mathbb R \to (0,1)$ defined by $f(x) =\frac 1 {\pi}(\arctan x +\frac {\pi} 2)$. This function is uniformly continuous (In fact Lipschitz, by MVT) and its inverse is continuous. [I am using the standard metrics on $\mathbb R$ and $(0,1)$]. $\mathbb R$ is complete and $(0,1)$ is not.