2
$\begingroup$

Prove that a function $f : X \mapsto X$ is injective if and only if it has a left inverse.

Can someone help me with this? What I've done is form predicates of the form $LI(f) \Rightarrow INJ(f)$ to begin step-by-step.

I understand the left hand side will be quantified to $\exists f,g: X \mapsto X: g \circ f = id_X$. Am I in the right direction here? Or am I missing something more that should be quantified? I know the following:

  1. the identity function, $id_X: X \mapsto X$, is defined by $\forall x \in X, id_X(x)=x$.
  2. A function $g \in \mathcal{F}$ is called a left inverse of a function $f \in \mathcal{F}$ if $g \circ f =id_X$.
$\endgroup$
  • $\begingroup$ The full quantification would be: $\forall f:X\to X, ((\forall x, y\in X, f(x) = f(y) \rightarrow x = y) \leftrightarrow (\exists g:X\to X, g \circ f = \mathrm{id}_X))$. $\endgroup$ – Daniel Schepler Jun 6 '17 at 22:33
  • $\begingroup$ @DanielSchepler What if we're doing it the other way around? Can we switch the statements then, obviously keeping/using $f$ and $g$ as required? $\endgroup$ – Yahya Farooq Jun 6 '17 at 22:37
  • $\begingroup$ It's unclear what you're asking. What do you mean by "doing it the other way around"? $\endgroup$ – Daniel Schepler Jun 6 '17 at 22:40
  • $\begingroup$ I mean, can it be expressed as $\forall X \mapsto X: g \circ f = id_X \Leftrightarrow \exists g: X \mapsto X,(\forall x,y \in X, f(x)=f(y) \Rightarrow x=y)$? $\endgroup$ – Yahya Farooq Jun 6 '17 at 22:43
  • $\begingroup$ That formula doesn't really make sense because the first occurrence of $g$ isn't bound by any quantifier. $\endgroup$ – Daniel Schepler Jun 6 '17 at 22:47
0
$\begingroup$

You can show injective implies a left inverse by constructing g.

First create a helper relation $ g' = \{ ( f(x), x ) : x \in X \}$ . This relation is a (partial) function due to the injective property ensuring that for every element in the domain there is at most one image.

Then you can construct $ g(x) = \begin{cases} g'(x) & \text{when } g'(x) \text{ is defined} \\ x & \text{otherwise} \end{cases} $

Which is a left inverse. You can show the converse by making use of the fact that for all functions $ x = y \rightarrow g(x) = g(y) $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.