# The closure of the graph of a certain composite in a topos.

Let $\mathcal{E}$ be an elementary topos with subobject classifier $\Omega$ and let $j\colon \Omega\to\Omega$ be a Lawvere-Tierney topology on it. Assume that, for an object $C$ of $\mathcal{E}$, each arrow $f\colon B\to C$ is such that (the subobject represented by) $G(f):= (1,f)\colon B\rightarrowtail B\times C$ is a closed subobject of $B\times C$. (I call $G(f)$ the graph of $f$). Take then any such $f\colon B\to C$ and suppose also given a monic arrow $m\colon A\rightarrowtail B$ which is dense (i.e. the subobject of $B$ represented by $m$ is dense in $B$, which means that its closure is the subobject represented by the identity on $B$).

Now, it is easy to see that $G(f\circ m)=(1,f\circ m)\colon A\to A\times C$ is the pullback along $m\times 1_{C}$ of $G(f)\colon B\rightarrowtail B\times C$. I also know that $m\times 1_{C}\colon A\times C\rightarrowtail B\times C$ is dense, as it is the pullback of $m$ along the projection of $B\times C$ into $B$. Using these facts, I am supposed to prove that the subobject represented by $$(m\times 1_{C})\circ G(f\circ m)=G(f)\circ m\colon A\rightarrowtail B\times C$$ (which is essentially $G(f\circ m)$ seen as a subobject of $B\times C$) has closure given by (the subobject represented by) $G(f)$. (This is a step in order to show that an object $C$ as above is separated, so I can not use this fact to show what I need to). I can not prove this assertion.

Up to now, I have just noticed that $G(f\circ m)$ is closed as a subobject of $A\times C$ and I have tried to compute the characteristic function of $G(f)\circ m$ hoping to show that $j\circ char(G(f)\circ m)=char (G(f))$, as this would precisely give what I am looking for. However, my attempt was unfruitful.

Any help would be appreciated. Thank you in advance.