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Stream: learning: questions

Topic: Do adjoints to projection maps form a functor?

Naso (May 04 2023 at 03:04):

Let $F : \mathcal{T}^{op} \to \mathcal{Pos}$ be a poset-valued presheaf on a topological space.

Suppose that all the projection maps $a \mapsto {a \mid_B}$ have a right adjoint, i.e. for $B \subseteq A$ and $b \in F B, a \in F A$,
${a \mid_B} \leq b \iff a \leq b^A$

Does it follow that $F^*$ defined by $F^*A = FA$ and $F^* (B \subseteq A) = b \mapsto b^A$ is a (covariant) functor?

I need to show that if $A \subseteq B \subseteq C$ then for each $a \in F^* A$, $a^C = (a^B)^C$.

I showed that $a^C \leq (a^B)^C$ using the Galois connection, functoriality of $F$ and the fact that left adjoint after right is an interior operator, but I am struggling to show the other direction.

Naso (May 04 2023 at 04:13):

Naso said:

Let $F : \mathcal{T}^{op} \to \mathcal{Pos}$ be a poset-valued presheaf on a topological space.

Suppose that all the projection maps $a \mapsto {a \mid_B}$ have a right adjoint, i.e. for $B \subseteq A$ and $b \in F B, a \in F A$,
${a \mid_B} \leq b \iff a \leq b^A$

Does it follow that $F^*$ defined by $F^*A = FA$ and $F^* (B \subseteq A) = b \mapsto b^A$ is a (covariant) functor?

I need to show that if $A \subseteq B \subseteq C$ then for each $a \in F^* A$, $a^C = (a^B)^C$.

I showed that $a^C \leq (a^B)^C$ using the Galois connection, functoriality of $F$ and the fact that left adjoint after right is an interior operator, but I am struggling to show the other direction.

Ok, maybe this is it?

By using the Galois connection twice and functoriality of projection, we get

$(a^B)^C \leq (a^B)^C \iff ((a^B)^C)_B)_A \leq a \iff ((a^B)^C)_A \leq a$

Then applying the Galois connection again we get

$(a^B)^C) \leq a^C$

Matteo Capucci (he/him) (May 04 2023 at 18:50):

Yeah these maps are still a functor

Matteo Capucci (he/him) (May 04 2023 at 18:52):

The best way to see this imo is to notice that adjunctions compose, hence if you have a functor picking out the left adjoints, the one picking out the right adjoints will be too

Naso (May 06 2023 at 01:51):

Matteo Capucci (he/him) said:

Yeah these maps are still a functor

Is there a name for this? Where for two functors $F : \mathcal{C^{op}} \to \mathcal{Cat}$ and $F^* : \mathcal{C} \to \mathcal{Cat}$, we have $F X= F^* X$ for all objects $X$ in $\mathcal{C}$ and $F f \dashv F^* f$ for all morphisms $f$ in $\mathcal{C}$?

Reid Barton (May 06 2023 at 05:58):

$F$ is a functor from $\mathcal{C}^{\mathrm{op}}$ to the category of (categories and adjunctions, viewed as morphisms in the direction of the left adjoint).

Reid Barton (May 06 2023 at 06:01):

In general we might want to insert some pseudo- and 2- prefixes in this statement, but because you said you were working with posets, it doesn't matter: as $a \le b \wedge b \le a \implies a = b$, there's no room for any invertible natural transformations between posets-viewed-as-categories.