A question about a practice problem

Question

Youtube has been helping me get used to working with sets and functions lately, and I've been following the lectures of Mathoma who has created an astounding set-theory playlist that goes through the elementary basics every student must be accustomed to. Now while discussing functions I found this practice problem:

Show that if $F$ is an injective function, that $F^{-1}$ is a function.

The first one requires us to prove the existence of the inverse function of $f$ using the premise stating that it is injective. I found this impossible for I have been taught that it must be true that a function must be both injective and surjective in order to have an inverse function. Perhaps, I said, instead of $f^{-1}$ what was meant to be written was 'left-inverse'?

Thank you in advance.

Answer 1

Suppose that $F:A \to B$ where $F$ is an injective function. Let $C$ denote the subset of $B$ represented by $F(A)$. That is, $C$ denotes the set that contains every element $c$ in $B$ such that there exists an element $a \in A$, where $F(a) = c.$

Then $F:A \to C$ is injective and surjective, and so there exists an inverse function $F^{-1}:C \to A$.

I suspect that that is the intended idea.

Answer 2

The problem comes from this video. Here, $\ F^{-1}\ $ denotes the inverse of the relation $\ F\ $, not necessarily of that the function $\ F\ $ for which some codomain has been specified.

The video gives the standard set theoretic definitions of relations and functions, a relation being a set of ordered pairs and a function being a special sort of relation—namely a relation $\ F\ $ such that $\ xFy_1\ \&\ $$xFy_2\Rightarrow$$\,y_1=y_2\ $. All relations have an inverse, defined in the segment from $9$:$19$ to $9$:$45$ of the video by $$ F^{-1}=\big\{\,\langle x,y\rangle\, \big|\, yFx\ \big\}\ . $$ What the exercise is asking you to do is prove that if $\ F\ $ is an injective function then the relation $\ F^{-1}\ $ is a function—that is that $$ xF^{-1}y_1\,\&\,xF^{-1}y_2\,\Rightarrow\,y_1=y_2\ . $$ The proof is pretty much a trivial one-liner.

If $\ F:B\rightarrow A\ $ is a function mapping $\ B\ $ into $\ A\ $, then the relation $\ F^{-1}\ $ (which always exists) isn't necessarily an inverse for that function, however. For this to be true, $\ F^{-1}\ $ must be a function with domain $\ A\ $. It will be a function if and only if $\ F\ $ is injective, and it's domain will be $\ A\ $ if and only if $\ A\ $ is the range of $\ F\ $—that is, if and only if $\ F\ $ is surjective. Nothing in the statement of the practice problem, nor in the rest of the video, as far as I can tell, contradicts this.