Galois Connection as morphism

Let $\mathbf{Gal}$$\mathbf{Gal}$ be a category such that:

• Objects are partially ordered sets

• For any two objects $(A,{\le }_A)$$(A,\leq_A)$ and $(B,{\le }_B)$$(B,\leq_B)$, morphisms $\text{Hom}(A,B)$$\text{Hom}(A,B)$ consist of pairs $(F,G)$$(F,G)$ where $F⊣G$$F \dashv G$ forms a Galois connection between $A$$A$ and $B$$B$

For $F⊣G:A\to B,H⊣K:B\to C$$F\dashv G:A\to B,H\dashv K:B\to C$, the composition $(H⊣K)\circ (F⊣G):=HF⊣GK:A\to C$$(H\dashv K)\circ (F\dashv G):=HF\dashv GK:A\to C$.

Well, we need to check that $HF$$HF$ and $GK$$GK$ do form a pair of Galois connection.

Consider $F:A\to B$$F:A\to B$ and $K:C\to B$$K:C\to B$. Then we have:

$$\begin{array}{}\text{(1)}& HF(a){\le }_{C}c⟺F(a){\le }_{B}K(c)⟺a{\le }_{A}GK(c)\end{array}$$

i.e.

$$\begin{array}{}\text{(2)}& HF(a){\le }_{C}c⟺a{\le }_{A}GK(c)\end{array}$$

Also, Easy to see that ${\mathrm{id}}_{\mathrm{A}}⊣{\mathrm{id}}_A$$\mathrm{id_A}\dashv\mathrm{id}_A$​ is a pair of Galois Connection, and the composition is associative.

Example. Let $f⊣f^{-1}:P(X)\to P(X)$$f\dashv f^{-1}:P(X)\to P(X)$, $\mathrm{Cl}⊣i:P(X)\to \mathrm{Clo}(X)$$\mathrm{Cl}\dashv i:P(X)\to \mathrm{Clo}(X)$ be two pairs of Galois connection.

Then consider the composition $(\mathrm{Cl}⊣i)\circ (f⊣f^{-1})=\mathrm{Cl}\circ f⊣f^{-1}\circ i$$(\mathrm{Cl}\dashv i)\circ(f\dashv f^{-1})=\mathrm{Cl}\circ f\dashv f^{-1}\circ i$. i.e.

$$\begin{array}{}\text{(3)}& \overline{f(A)}\subseteq B⟺A\subseteq f^{-1}(B)\end{array}$$

Let us define a functor on $\mathbf{Gal}$$\mathbf{Gal}$, $\mathcal{D}:{\mathbf{Gal}}^{\mathbf{op}}\to \mathbf{Gal}$$\mathcal{D}:\mathbf{Gal^{op}}\to\mathbf{Gal}$​, by:

• For object $(A,\le )$$(A,\leq)$, $\mathcal{D}(A,\le )=(A,\ge )$$\mathcal{D}(A,\leq)=(A,\geq)$ (i.e., $A^{\mathrm{op}}$$A^{\mathrm{op}}$)

• For morphism $(F,G):A\to B$$(F,G):A\to B$, $\mathcal{D}(F,G)=(G,F):B^{\mathrm{op}}\to A^{\mathrm{op}}$$\mathcal{D}(F,G)=(G,F):B^{\mathrm{op}}\to A^{\mathrm{op}}$

This is an isomorphism ${\mathbf{Gal}}^{\mathbf{op}}\cong \mathbf{Gal}.$$\mathbf{Gal^{op}}\cong\mathbf{Gal}.$ The inverse of $\mathcal{D}$$\mathcal D$ is itself.

If we consider a full subcategory of $\mathbf{Gal}$$\mathbf{Gal}$ such that the object satisfy ${\eta }_A:A\cong A^{\mathrm{op}}$$\eta_A:A\cong A^{\mathrm{op}}$, denote it as ${\mathbf{Gal}}_{†}$$\mathbf{Gal}_{\dagger}$​.

Let us define the functor $†:{\mathbf{Gal}}_{†}^{\mathbf{op}}\to {\mathbf{Gal}}_{†}$$\dagger: \mathbf{Gal_{\dagger}^{op}}\to\mathbf{Gal}_{\dagger}$ via this natural isomorphism:

$$\begin{array}{}\text{(4)}& \begin{array}{c}\begin{array}{ccc}X& \stackrel{{\eta }_X}{\to }& X^{\mathrm{op}}\\ (F,G{)}^{†}\downarrow & & \downarrow \mathcal{D}(F,G)& \\ Y& \stackrel{{\eta }_Y}{\to }& Y^{\mathrm{op}}\end{array}\end{array}\end{array}$$

Then it could be a dagger category.

Remark. This dagger construction is inspired by the relationship between adjoint operators and dual maps in linear algebra.