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In my topology notes the definition is given as:

A function $f : X \rightarrow Y$ is said to be a homeomorphism if: $f$ is continuous, bijective, and moreover its inverse $f^{−1} : Y \rightarrow X$ is continuous.

My question is: Does the fact that $f$ has an inverse not already guarantee that $f$ is bijective?

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    $\begingroup$ It guarantees its inverse, which does not mean it is necessarily continuous. $\endgroup$ – DrHAL Feb 9 '16 at 19:06
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    $\begingroup$ If you'd define "$f$ is continuous and its inverse ..." your audience would interrupt you and shout "What inverse? Do you mean $f$ is in fact bijective? Or injective and you mean left inverse? Or surjective and you mean right inverse? But those are not even unique in general! Help, I'm lost!". You might shorten to "$f$ is continuous and has a continuous two-sided inverse". Or, more leaning towards category theory: "$f$ is continuous and there exists continuous $g\colon Y\to X$ with $f\circ g=\operatorname{id}_Y$ and $g\circ f=\operatorname{id}_X$." $\endgroup$ – Hagen von Eitzen Feb 9 '16 at 19:08
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The fact that $f$ is bijective ensures that $f^{-1}$ is a valid function from $Y$ to $X$. We now want both $f$ and $f^{-1}$ to be continuous, and the latter is not automatic from the former (consider the identity between $X$ in the discrete topology to $X$ in a non-discrete topology, e.g.). So we demand them separately.

We could just say $f$ and $f^{-1}$ are both continuous, but then we'd be implicitly assuming the bijectivity of $f$ by just "using" a function $f^{-1}$. It's better to be explicit here, and state the bijectiveness separately. Hence the order: $f$ continuous, bijective (so that $f^{-1}$ is a valid function!) and finally this function $f^{-1}: Y \rightarrow X$ is also continuous.

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