Equivalence of Definitions of Injection/Definition 1 iff Definition 4

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

Let $$f^{-1}$$ be the inverse of $$f$$.

Let $$f^{-1} \left({t}\right)$$ be the preimage of $$t \in T$$.

Then $$f^{-1} \left({t}\right)$$ is a singleton for all $$t \in T$$ iff $$f$$ is an injection.

Proof
Follows immediately from the definition of injection.


 * Let $$\exists t \in T: s_1, s_2 \in f^{-1} \left({t}\right), s_1 \ne s_2$$.

Then we have:
 * $$f \left({s_1}\right) = f \left({s_2}\right)$$ but $$s_1 \ne s_2$$

and so $$f$$ is not an injection.

So, by the Rule of Transposition, if $$f$$ is an injection then $$f^{-1} \left({t}\right)$$ has no more than one element.


 * Suppose $$f$$ is not an injection.

Then by definition:
 * $$\exists s_1, s_2 \in S, s_1 \ne s_2: f \left({s_1}\right) = f \left({s_2}\right) = t$$

By definition of preimage of $$t \in T$$:
 * $$s_1 \in f^{-1} \left({t}\right), s_2 \in f^{-1} \left({t}\right)$$

and so: $$\left\{{s_1, s_2}\right\} \subseteq f^{-1} \left({t}\right)$$

So if $$f$$ is not an injection then $$f^{-1} \left({t}\right)$$ has more than one element for at least one $$t \in T$$.

Hence the result.