j-closed monomorphisms in a topos with a Lawvere-Tierney topology j

Question

Let’s assume $\epsilon$ is a topos and j is a Lawvere-Tierney topology in $\epsilon$. Now, let $Sh_j(\epsilon)$ be the full subcategory of $\epsilon$ on the sheaves for j. Now, let $i:Sh_j(\epsilon)\longrightarrow \epsilon$ be the embedding functor and let $L:\epsilon\longrightarrow Sh_j(\epsilon)$ be the sheafification functor. Then, $L\dashv i$. $Sh_j(\epsilon)$ is a topos and let $\Omega_j$ be its subobject classifier and $1_j$ be its terminal object. Now, let $m:X\longrightarrow Y$ be a j-closed monomorphism. How can we show that there exists a unique morphism $\phi:Y\longrightarrow \Omega_j$ which makes the square with vertices X,Y, $i(1_j)$ and $i(\Omega_j)$ a pullback.

Answer 1

I will work entirely in the topos $\epsilon$. Since $i$ preserves limits we have that $i(1_j)$ is just the terminal object in $\epsilon$.

The object $\Omega_j$ can be found as a subobject of $\Omega$ via the equalizer: $$ \Omega_j \xrightarrow{\omega_j} \Omega \overset{j}{\underset{Id}{\rightrightarrows}} \Omega $$ See for example Sheaves in Geometry and Logic by Mac Lane and Moerdijk ((7) on page 224).

Let $\chi: Y \to \Omega$ classify $m: X \to Y$. Since $m$ is closed we have $jm = m$. So by the universal property of the equalizer, there is $\phi: Y \to \Omega_j$ such that $\omega_j \phi = \chi$. I claim that this $\phi$ is the $\phi$ you asked for.

Since $t: 1 \to \Omega$ satisfies $jt = t$, we can again use the universal property of the equalizer to find $t_j: 1 \to \Omega_j$ such that $\omega_j t_j = t$.

We can thus form the following commuting diagram. $\require{AMScd}$ \begin{CD} X @>>> 1 @= 1\\ @V m VV @V t_j VV @VVtV\\ Y @>> \phi > \Omega_j @>> \omega_j > \Omega \end{CD} Because $\chi$ classifies $m$ and the bottom arrow is just $\chi$, the outer rectangle is a pullback.

To see that the left square is a pullback, we check the universal property. Let $f: Z \to Y$ and $g: Z \to 1$ be such that $\phi f = t_j g$. Then $\omega_j \phi f = \omega_j t_j g = t g$. So there is unique $u: Z \to X$ making everything commute.

We are left to show that $\phi$ is unique. Suppose that we have $\psi: Y \to \Omega_j$ such that the left square below is a pullback: \begin{CD} X @>>> 1 @= 1\\ @V m VV @V t_j VV @VVtV\\ Y @>> \psi > \Omega_j @>> \omega_j > \Omega \end{CD} Then the outer rectangle is also a pullback. To see this let $f: Z \to Y$ and $g: Z \to 1$ be such that $\omega_j \psi f = t g$. Then since $t g = \omega_j t_j g$ and $\omega_j$ is mono, we have $\psi f = t_j g$. We assumed that the left square is a pullback, so there we find the required unique arrow $Z \to X$. Since the outer square is a pullback we have that $\omega_j \psi$ classifies $m$. Since classifiers are unique, we have $\omega_j \psi = \chi = \omega_j \phi$. Then by how we constructed $\phi$ from the equalizer, we conclude that $\psi = \phi$.