Injection iff Inverse of Image Mapping

Theorem

A mapping $f$ is an injection iff:

$f^{-1}: \operatorname{Im} \left({f}\right) \to \operatorname{Dom} \left({f}\right)$

is a mapping.

Proof

Necessary Condition

Let $f: S \to T$ be an injection.

First we note that:

$t \in \operatorname{Im} \left({f}\right) \implies \exists x \in \operatorname{Dom} \left({f}\right): f \left({x}\right) = t$

thus fulfilling the condition $\forall y \in T: \exists x \in S: f \left({x}\right)= y$.

Now let $t \in \operatorname{Im} \left({f}\right): \left({t, y}\right), \left({t, z}\right) \in f^{-1}$.

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({t, y}\right), \left({t, z}\right)\)	\(\in\)	\(\displaystyle f^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({y, t}\right), \left({z, t}\right)\)	\(\in\)	\(\displaystyle f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Inverse
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle f \left({y}\right) = t\)	\(=\)	\(\displaystyle f \left({z}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Equality of Elements in Range of Mapping
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y\)	\(=\)	\(\displaystyle z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $f$ is injective

So by the definition of mapping, $f^{-1}$ is a mapping.

$\Box$

Sufficient Condition

Let $f^{-1}$ be a mapping.

We need to show that $\forall x, z \in \operatorname{Dom} \left({f}\right): f \left({x}\right) = f \left({z}\right) \implies x = z$.

So, pick any $x, z \in \operatorname{Dom} \left({f}\right)$ such that $ f \left({x}\right) = f \left({z}\right)$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f \left({x}\right)\)	\(=\)	\(\displaystyle f \left({z}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \exists y \in \operatorname{Dom} \left({f}\right): \left({x, y}\right), \left({z, y}\right)\)	\(\in\)	\(\displaystyle f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of a Mapping
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({y, x}\right), \left({y, z}\right)\)	\(\in\)	\(\displaystyle f^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Inverse
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as it is specified that $f^{-1}$ is a Definition of a Mapping

Thus by the definition of an injection, $f$ is an injection.

$\blacksquare$

Comment

Some sources, for example W.E. Deskins: Abstract Algebra (1964), use this property as the definition of an injection.

Sources

• Nathan Jacobson: Lectures in Abstract Algebra: I. Basic Concepts (1951): Introduction $\S 2$ (passim)
• W.E. Deskins: Abstract Algebra (1964): $\S 1.3$: Definition $1.9 \ \text{(a)}$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 5$: Theorem $5.2$
• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Notation and Terminology