Mapping is Injection iff Direct Image Mapping is Injection

Theorem

Let $f: S \to T$ be a mapping.

Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.

Then:

$f^\to$ be an injection

if and only if

$f: S \to T$ is also an injection.

Proof

Necessary Condition

Suppose $f: S \to T$ is a mapping, but not injective.

Then:

$\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2} = y$

Let:

$X_1 = \set {x_1}$

$X_2 = \set {x_2}$

$Y = \set y$

Then it follows that:

$\map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$

Thus $f^\to: \powerset S \to \powerset T$ is not injective.

So by the Rule of Transposition, the result follows.

$\Box$

Sufficient Condition

Suppose $f^\to: \powerset S \to \powerset T$ is not an injection.

Then:

$\exists Y \in \powerset T: \exists X_1, X_2 \in \powerset S: X_1 \ne X_2 \land \map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$

There are two cases to consider:

$(1): \quad$ Either $X_1$ or $X_2$ is the empty set

$(2): \quad$ Neither $X_1$ nor $X_2$ is the empty set.

Without loss of generality assume that $X_1 = \O$.

Then, from Image of Empty Set is Empty Set:

$Y = \map {f^\to} {X_1} = \O$

But that means:

$\map {f^\to} {X_2} = \O$

Now $X_1 \ne X_2$, so $X_2 \ne \O$.

Thus:

$\exists x_2 \in X_2: \neg \exists y \in T: \map f {x_2} = y$

which means that $f$ is not even a mapping, let alone an injection.

The same argument applies if $X_2$ is empty.

$\Box$

Now, assume that neither $X_1$ nor $X_2$ is the empty set.

As $X_1 \ne X_2$, there must be at least one element in either one which is not in the other.

Without loss of generality assume that $\exists x_1 \in X_1: x_1 \notin X_2$.

Now suppose $\map f {x_1} = y \in Y$.

However, as $\map {f^\to} {X_2} = Y$, there must be some $x_2 \in X_2$ such that $\map f {x_2} = y \in Y$.

So:

$\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2}$

which means, by definition, that $f$ is not injective.

Thus, if $f^\to: \powerset S \to \powerset T$ is not an injection then neither is $f: S \to T$.

Thus, by the Rule of Transposition, the result follows.

$\blacksquare$