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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.

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

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