Preorder as a monoid object

Consider the category of endorelations on a set $X$$X$. Specifically, let $P(X\times X)$$P(X \times X)$ form a partial order category where the objects are elements of $P(X\times X)$$P(X \times X)$ (i.e., relations on $X$$X$), and the morphisms are inclusions. We denote this category as $\underset{―}{X}$$\underline{X}$.

This category admits a natural monoidal structure. The monoidal product is defined as relational composition, with the unit element being the diagonal relation $\mathrm{\Delta }$$\Delta$.

We observe that a preorder on $X$$X$ is precisely a monoid object in $(\underset{―}{X},\circ ,\mathrm{\Delta })$$(\underline{X}, \circ, \Delta)$.

The reason is straightforward: a relation $R$$R$ is transitive if and only if $R\circ R\subseteq R$$R \circ R \subseteq R$, and a relation is reflexive if and only if $\mathrm{\Delta }\subseteq R$$\Delta \subseteq R$.

The remaining associativity and unitality laws are evident. Since we are working in a partial order category, any diagram one can draw is commutative, and this forms a monoidal structure.

In addition, we have

$$\mathsf{Sym}(X)=\mathsf{Pic}(\underset{―}{X})$$

 

Socle as Right adjoint Functor.

Let $k$$k$ be a field and let $A$$A$ be a $k$$k$-algebra. Define

$$\mathsf{Rep}(A):=\mathsf{Add}(B(A),{\mathsf{Vect}}_k),$$

which is an abelian category.

Definition.

• A representation $V$$V$ is simple if $\mathsf{Sub}(V)\cong \{0,1\}$$\mathsf{Sub}(V)\cong\{0,1\}$.

• A representation $V$$V$ is semisimple if

  $$V=\underset{i\in I}{⨁}{S}_i,\text{where each }{S}_i\text{ is simple.}$$

  Let $\mathsf{Semi}(A)$$\mathsf{Semi}(A)$ be the full subcategory of semisimple representations. This is an abelian category.

Proposition. $\mathsf{Semi}(A)$$\mathsf{Semi}(A)$ is a coreflective subcategory of $\mathsf{Rep}(A)$$\mathsf{Rep}(A)$.

We define the right adjoint of the inclusion $i:\mathsf{Semi}(A)\hookrightarrow \mathsf{Rep}(A)$$i:\mathsf{Semi}(A)\hookrightarrow\mathsf{Rep}(A)$ as follows.

For $V\in \mathsf{Rep}(A)$$V\in\mathsf{Rep}(A)$, set

$$\mathrm{soc}(V):=\text{the sum of all simple subrepresentations of }V.$$

This assignment defines a functor because for every morphism $f:V\to W$$f:V\to W$ we have

$$f(\mathrm{soc}(V))\subseteq \mathrm{soc}(W).$$

Moreover, $\mathrm{soc}(-)$$\operatorname{soc}(-)$ is additive.

The adjunction can be written as: for every $S\in \mathsf{Semi}(A)$$S\in\mathsf{Semi}(A)$ and every $V\in \mathsf{Rep}(A)$$V\in\mathsf{Rep}(A)$,

$${\mathrm{Hom}}_{\mathsf{Rep}(A)}(i(S),V)\cong {\mathrm{Hom}}_{\mathsf{Semi}(A)}(S,\mathrm{soc}(V)),$$

naturally in $S$$S$ and $V$$V$. Hence $\mathrm{soc}:\mathsf{Rep}(A)\to \mathsf{Semi}(A)$$\operatorname{soc}:\mathsf{Rep}(A)\to\mathsf{Semi}(A)$ is right adjoint to the inclusion $i$$i$.

Corollary. $\mathrm{soc}(-)$$\operatorname{soc}(-)$ is left exact.