Thm. $g:X \rightarrow Y$ is invertible iff $g$ is bijective.

Pf. Two implications need to be proven:
1. $g$ is invertible $\rightarrow$ $g$ is bijective.
2. $g$ is bijective $\rightarrow$ $g$ is invertible.
ad (1): The proposition has it that $g$ is invertible $\leftrightarrow (\exists h:Y\rightarrow X) \: g \circ h = \text{id}_{Y} \wedge h \circ g = \text{id}_{X}$. Or, equivalently, $$(\exists h)(\underbrace{(\forall y \in Y) \: g(h(y)) = y}_{(i)} \: \wedge \: \underbrace{(\forall x \in X) \: h(g(x)) = x)}_{(ii)}.$$ I.e. assume, according to (i), there is a function $h$ such that for every element in $Y$ there is an element $h(y) = x \in X$ such that $g(x) = y$. This is the definition for $g$ to be onto/surjective. Now assume that $g$ is not 1-1: $$(\exists x,x' \in X) \: x \neq x' \wedge g(x) = g(x').$$ Since $h$ is a function this means that $$(\forall g(x),g(x') \in Y) \: g(x) = g(x') \rightarrow h(g(x)) = h(g(x')).$$ So we can deduce $$g \text{ is not onto } \rightarrow (\exists x,x' \in X) \: ( h(g(x')) = h(g(x)) = x' \neq x ) \leftrightarrow (\forall h)(\exists x \in X) \: h(g(x)) \neq x,$$ where the contrapositive states that $$(\exists h)(\forall x \in X) \: h(g(x)) = x \rightarrow g \text{ is onto}.$$ Hence, (ii) implies $g$ to be an injection.
2015 MAR 19 (v1.0)
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