State/Injective functions have left inverses

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Proposition
Proof

Proposition: (Injective functions have left inverses) Suppose that $X$ and $Y$ are sets, and $f$ is a function from $X$ to $Y$. Then, $f$ is injective if and only if there exists a function $g \colon Y \rightarrow X$ such that $g \circ f = Id_X$, i.e., a left inverse of $f$.

Logical Connections

This statement logically relies on the following definitions and statements: Def/Function, Def/Injective, Def/Inverse function, Def/Image, Def/Extension of functions, Def/Complement of a set

The following statements and definitions rely on the material of this page: State/Bijections are injective and surjective functions, and State/Composing injective surjective or bijective functions yields the same

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State/Injective functions have left inverses

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	Suppose first that $f$ is injective.
	Let $I$ be the image of $f$. If $i \in I$, then $\exists ! x \in X$ such that $f(x) = i$. Thus, there exists a unique function $g \colon I \rightarrow X$ such that $g(i) = x \Leftrightarrow f(x) = i$. In particular, $g(f(x)) = x$, for all $x \in X$. Now, we may extend $g$ to a function from $Y$ to $X$, by choosing any element $x_0 \in X$, and defining $g(y) = x_0$ for all $y \in Y - I$ (the complement of $I$ in $Y$). Then, it follows that $g(f(x)) = x$ for all $x \in X$, since $f(x) \in I = Im(f)$ for all $x \in X$.
	Thus $g \circ f = Id_X$.

	Suppose that $f$ has a left inverse, $g$, i.e., that $g \circ f = Id_X$.
	Suppose that $x_1, x_2 \in X$, and $f(x_1) = f(x_2)$. Then, we find that $$x_1 = g(f(x_1)) = g(f(x_2)) = x_2.$$
	Hence $f$ is injective.