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Def 1: A function $f$ is said to be injective just in case: $$\forall a, b \in \mathrm{dom}(f): f(a) = f(b)\Longrightarrow a = b.$$

Will this alternative definition capture the same idea?

Def 2: A function $f$ is said to be injective just in case: $$f^{-1}~\rm{is~a ~function}.$$

If 'no', what might be added to make it work?

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up vote 6 down vote accepted

Yes, as long as you pick the right domain for the inverse, otherwise it is not even defined. The domain has to be the image of $f$ or a subset of it for the inverse to be defined.

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Assuming $f^{-1}$ is defined on the appropriate domain, you can think of it like this:

$f^{-1}$ is a function $\Rightarrow \forall \hat a,\hat b \in \mathrm {dom} f^{-1}, \hat a = \hat b \Rightarrow f^{-1}(\hat a) = f^{-1}(\hat b)$

but $\hat a = f(a), \hat b = f(b)$, where $a, b \in \mathrm {dom} f$

and $f^{-1}(\hat a) = f^{-1}(f(a)) = a, f^{-1}(\hat b) = f^{-1}(f(b)) = b$.

So we have $f(a) = f(b) \Rightarrow a = b$.

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