# Uniform Isomorphism

I have read somewhere that a function $$f:(X,\mathcal{D}(X))\longrightarrow (Y,\mathcal{D}(Y))$$ is a uniform isomorphism provided that $$f$$ and $$f^{-1}$$ are uniform continuous functions and $$f$$ is bijective.

But, can we really find a uniformly continuous function $$f$$ such that $$f^{-1}$$ is uniformly continuous, but $$f$$ is not bijective?

I am asking this because I look at $$f^{-1}$$ as a function that supposed to move from $$Y$$ to $$X$$, and for such function to exist, $$f$$ must be bijective.

• if $f^{-1}$ is a function, $f$ must be bijective. – YuiTo Cheng Mar 24 at 8:34

Of course, $$f^{-1}$$ as a function $$Y \to X$$ exists iff $$f$$ is bijective; this is standard set theory.
So saying that $$f^{-1}$$ is uniformly continuous implies $$f^{-1}$$ exists as a function and hence $$f$$ (and $$f^{-1}$$) is bijective.
I suppose the extra condition of the bijectivity of $$f$$ is just to ensure the existence of the $$f^{-1}$$ that you want to assert the uniform continuity of.