Proof involving functional graphs and the injective property

[Uncanny]

TL;DR Summary

The problem I’ve attempted is number 12 in the attached thumbnails. Also included a bit above the problem statement is the relevant definition (i.e. that of a functional graph). My proof is also attached.

My only qualm is that the statement “Let G be a functional graph” never came into play in my proof, although I believe it to be otherwise consistent. Can someone take a look and let me know if I missed something, please? Or is there another reason to include that piece of information?

 

[Uncanny]

Here’s an update with things written out:

Definition:

Let $G$ be a graph. $G$ is a functional graph if and only if $(x_1,y_1) \in G$ and $(x_1,y_2) \in G$ implies $y_1=y_2$.

Problem statement, as written:

Let $G$ be a functional graph. Prove that $G$ is injective if and only if for arbitrary graphs $J$ and $H$: $G \circ (H \cap J) = (G \circ H) \cap (G \circ J).$

I am concerned with the direction of the biconditional proving that $G$ is injective. I have completed a proof which seems consistent to me, but does not utilize the premise of $G$ being a functional graph anywhere in the argument (including the other direction of the biconditional). Is it simply necessary, a priori, for a graph to be a functional graph in order for it to be considered injective?

I can post my proof if needed, but here is the gist: I suppose the antecedent (assume for arbitrary graphs $J,H$ that the equality written above holds). Then I proceed via contradiction, supposing (to the contrary) $G$ is not injective. This brings into existence two elements (ordered pairs) of $G$ with unequal first elements, but equal second elements. From here, I construct $J,H$ such that I may use the equality given above to show that the unequal first elements, in fact, are equal- a contradiction.

*Nowhere do I use the fact that $G$ is a functional graph. For some reason, this isn’t sitting well. Is my conjecture above true about graphs unable to be considered injective if they are not first classified as functional?