A category admitting pullbacks and pullback stability

Question

Let's say $C$ is a category, and $\mathscr{C}$ is a collection of morphisms in $C$. I have come across the following sentence

"$C$ admits pullbacks along morphisms from $\mathscr{C}$, and $\mathscr{C}$ is pullback stable"

I know what a pullback is, but have no idea what this sentence means. Here are my attempts.

1. $C$ admits pullbacks along morphisms from $\mathscr{C}$: Suppose $(P,p,q)$ is a pullback of $f$ and $g$, then $f$ and $g$ are in $\mathscr{C}$.
2. $\mathscr{C}$ is pullback stable: Suppose $(P,p,q)$ is a pullback of $f$ and $g$, then if $g$ has a certain property so does $p$. But which property are we talking about here? Is it talking about belonging to $\mathscr{C}$?

Any correction or help will be greatly appreciated.

Answer 1

Saying that $C$ admits pullbacks along arrows in $\mathscr{C}$ means that whenever we have arrow $A \xrightarrow{f} C \xleftarrow{g} B$, with at least one of them in $\mathscr{C}$, we can form their pullback. For example, if $f$ is in $\mathscr{C}$, then we consider this the pullback of $g$ along $f$.

If we say that $\mathscr{C}$ is stable under pullbacks, then we mean that if we pullback an arrow in $\mathscr{C}$ along another arrow, the resulting arrow is in $\mathscr{C}$ again. That is, if we have the following pullback diagram: $\require{AMScd}$ \begin{CD} D @>f'>> B\\ @Vg'VV @VVgV\\ A @>>f> C \end{CD} If $f$ is in $\mathscr{C}$, then $f'$ should also be in $\mathscr{C}$.

Answer 2

In general, I would interpret this sentence as saying that if $f:X\to Z$ is any morphism in the category $C$ and $g:Y\to Z$ is a morphism in the class $\mathcal{C}$, then they have a pullback $$\require{AMScd} \begin{CD} P@>{p}>> Y \\ @V{q}VV @VV{g}V \\ X@>>{f}>Z \end{CD}$$ and $q$ is in the class $\mathcal{C}$ (so that part is as you said). Note that technically the pullback is only defined up to isomorphism, so if your class $\mathcal{C}$ is not closed under (pre)composition with isomorphisms this could be ambiguous, but this is relatively rare (and it shouldn't happen for Galois structures).

Answer 3

The following is more or less the same as the other answers but is packaged in a different manner that I, personally, find more convenient/understandable.

Given any category $\mathcal E$, we have the arrow category $\mathcal E^{\to}$ whose objects are arrows of $\mathcal E$ and whose arrows are pairs of arrows in $\mathcal E$. For example, an arrow from $f:A\to B$ to $g:C\to D$ would be a pair of arrows $h:A\to C$ and $k:B\to D$ such that $k\circ f = g\circ h$.

If $\mathcal E$ has pullbacks, then the codomain functor $\mathsf{cod}:\mathcal E^{\to}\to\mathcal E$ turns $\mathcal E^{\to}$ into a fibred category called the codomain fibration. For each object $A\in\mathcal E$, we have a fiber category $\mathcal E_A$ consisting of the subcategory of $\mathcal E^{\to}$ (called the "total category") whose objects get sent to $A$ by $\mathsf{cod}$ and whose arrows get sent to $id_A$. In this case, this corresponds exactly to the slice category $\mathcal E/A$. For any arrow, $f:A\to B$, in the base category, which is $\mathcal E$ in this case, we have a functor $f^*:\mathcal E_B\to\mathcal E_A$ defined by pulling back an object of $\mathcal E_B$ (which is an arrow into $B$) along $f$ which will produce an arrow into $A$ (and thus an object of $\mathcal E_A$) as one of the projections of the pullback. Often a property is considered pullback stable if it is preserved by these pullback functors.

For your scenario, instead of considering all of $\mathcal E^{\to}$, we can consider the full subcategory consisting of the arrows (viewed as objects) in $\mathscr C$. Now we don't need $\mathcal E$ to have all pullbacks. If we start to define a restricted codomain fibration, then we see need pullbacks of arrows in $\mathscr C$ along arrows of $\mathcal E$ need to exist so we can define the pullback functors, this is what admitting pullbacks along morphisms in $\mathscr C$ gives us. We also need the arrows produced by these pullback functors to be in $\mathscr C$ or else the functors simply won't land in our (restricted) fiber categories. This is what pullback stability gives us.

In other words, the constraints are more or less equivalent to saying that the codomain functor $\mathcal E^{\to}\to\mathcal E$ when restricted to the full subcategory of $\mathcal E^{\to}$ consisting of the arrows in $\mathscr C$ is a fibration.