Only bijective mappings are invertible. Clarifying proof. – math.stackexchange.com

Only bijective mappings are invertible. Clarifying proof. – math.stackexchange.com

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Suppose that $f : A \mapsto B$ is invertible with inverse $g : B \mapsto A$. Then $g \circ f = \operatorname{id}_A$ which means $\forall a \in A : g(f(a)) = a$. Let's take some $a_1, a_2 \in A$ with ...

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