I Proof involving functional graphs and the injective property

[Uncanny]

TL;DR Summary

I retyped a question I previously posted incase reading my handwriting was keeping some from interacting:

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?

 

[Uncanny]

Ah shoot.. sorry guys. Figuring out the formatting

 

[andrewkirk]

I presume, although it is not stated here, that H and J are also required to be functional graphs, and to have the same vertex set as G.

The result, in this direction at least, appears to be true if we replace 'functional graph' everywhere by 'function'. So you're correct that it doesn't use the notion of functional graph as distinct from a function.

PS You need to use a double-# instead of $ for in-line latex on this site.

 

[Uncanny]

thanks, I fixed that.

Thanks for your reply. $J$ and $H$ are actually only given to be arbitrary graphs. I attached a photo of the page the problem appears on (I’m working out of Charles Pinter’s “Book of Set Theory”) so you can verify this.

I have a feeling the premise “G must be a functional graph” is included because of some technicality (which I’m not familiar with) involving the injective property applied to graphs which must be functional a priori.

Nonetheless, it seems one can construct a graph that can satisfy the injective property without being a functional graph [$(x,y),(x,z) \in G$ still technically satisfies the injective property, although $G$ is clearly not functional].