Subobject classifier of a general slice category

category-theory

Let $\bf C$ be a category with finite limits, with subobject classifier $\omega:T\to \Omega$, where $T$ is the terminal object of $\bf C$. I'll denote $(X,x)$, with $x:X\to A$, an object in the slice category $\mathbf{C}/A$ (over any object $A$ in $\bf C$) . How can I describe the subobject classifier of $ \mathbf{ C}/A$?

I thought of $\tau:(A,1_A)\to(A\times \Omega,\pi_A)$ defined by $\pi_\Omega\circ \tau=\omega \circ t_A$, where $t_A$ is the unique $A\to T$, and $\pi_A\circ \tau=1_A$. In this way for every monomorphism $m:(S,s)\to(X,x)$, we can use the characteristic function $\chi_S:X\to \Omega$ to obtain this commutative diagram, setting $\pi_A\circ \chi_{(S,s)}=x$ and $\pi_\Omega\circ \chi_{(S,s)}=\chi_S$.

I have doubts about the fact that it is a pullback though. Let's take another monomorphism $m':(S',s')\to (X,x)$ such that $\tau\circ s'=\chi_{(S,s)}\circ m'$ (diagram below). $\omega\circ (s'\circ t_A)=\chi_S\circ m'$, and $m\circ f=m'$ for an $f:S'\to S$. Moreover one can prove that this $f$ is an arrow $(S',s')\to (S,s)$.

Finally, since $m$ is a monomorphism, I'd say that the uniqueness of $f$ follows from $m\circ f=m'$.

Do my arguments make sense? I have not checked explicitely the commutativity of all triangles/squares necessary, but it should be quite immediate. I don't know if it is legit to use pics instead of latex (let me know if it isn't); I really couldn't understand how to draw those diagrams.

Best Answer

A subobject in $\mathbf C/X$ consists of a monomorphism $U\hookrightarrow Y$ and a morphism $Y\to X$. If you have a subobject classifier $\top\hookrightarrow\Omega$, then $U\hookrightarrow X$ is the pullback of $\top\hookrightarrow\Omega$ along a unique morphism $Y\to\Omega$. If you have products, the fixed morphism $Y\to X$ gives a correspondence of morphisms $Y\to\Omega$ to morphism $Y\to X\times\Omega$ such that $Y\to X$ factors as $Y\to X\times\Omega\to X$, in which case $Y\to\Omega$ factors as $Y\to X\times\Omega\to\Omega$. By the pullback lemma, pullbacks of $\top\hookrightarrow\Omega$ along $Y\to\Omega$ correspond to pullbacks along $Y\to X\times\Omega$ of the pullback $X\times\top\hookrightarrow X\times\Omega$ of $\top\hookrightarrow\Omega$ along the projection $X\times\Omega\to\Omega$.