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 $\psi = \phi$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.