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This is probably a pretty general question and might require a little more information, but suppose that a function with a defined Domain and Co-Domain has an inverse function whose Domain is the Co-Domain of the original function and whose Co-Domain is the Domain of the original function.

Is this the same thing as the function being bijective? The reason I ask is that I've noticed a lot of the proofs in my Algebra class that involve showing a function is a bijection consist of only defining an inverse function and nothing more.

Oderus
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    A function from $X$ to $Y$ is one-to-one if and only if it has a left inverse; it is surjective if and only if it has a right inverse; and it is bijective if and only if it has a (two-sided) inverse. This is a theorem. – Arturo Magidin Oct 16 '20 at 01:21
  • What do you mean by left and right inverses? – Oderus Oct 16 '20 at 03:15
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    If $f\colon X\to Y$, then a function $g\colon Y\to X$ is (i) a left inverse of $f$ if and only if $g\circ f = \mathrm{id}_X$; (ii) a right inverse of $f$ if and only if $f\circ g=\mathrm{id}_Y$; and (iii) a (two-sided) inverse of $f$ if and only if $f\circ g=\mathrm{id}_Y$ and $g\circ f=\mathrm{id}_X$, if and only if it is both a left and a right inverse of $f$. – Arturo Magidin Oct 16 '20 at 03:25
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    See here, – Arturo Magidin Oct 16 '20 at 03:38

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