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

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.

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 $$\psi = \phi$$.