Zoran Skoda subfunctor of identity

Let $A$ be a concrete category.

An endofunctor $F : A\to A$ is a subfunctor of the identity functor if $F(M)\subset M$ for each object $M$ in $A$. That means in particular that the inclusion is a natural transformation. Thus, if $g:M\to N$ is a morphism in $A$, $F(g)$ is simply the restriction-corestriction of $g$ and $g(F(M)) = F(g)(M)\subset F(N)$.

How to make sense of this in an arbitrary category. Subobject is viewed as an isomorphism class of monomorphisms. So a subfunctor of the identity should be a pair $(F,i)$ where $F$ is an endofunctor and $i:F\Rightarrow Id_A$ is the natural transformations whose components $i_M:F(M)\hookrightarrow M$ are monic. The commutative diagram

$$\begin{array}{ccc} F(M) & \overset{i_M}\hookrightarrow & M\\ F(g)\downarrow & &\downarrow g\\ F(N) & \overset{i_N}\hookrightarrow & N\\ \end{array}$$

defines the restriction $g\circ i_M:F(M)\to N$ of $G$ and then, by monicity of $i_N$, determines the arrow $F(M)\to F(N)$ uniquely; this is the restriction-corestriction when it exists.

In the particular case if $g$ is also an inclusion, $L\subset N$. It is clear that $F(L)\subset L\cap F(N)$

If $g = i_M$ then by monicity of $i_M$ from the following commutative diagram

$$\begin{array}{ccc} F F M & \overset{i_{F M}}\hookrightarrow & F M\\ F(i_M)\downarrow & &\downarrow i_M\\ F M & \overset{i_M}\hookrightarrow & M\\ \end{array}$$

we obtain $F(i_M) = (i_M)|_{F(M)} = i_{F M}$, that is, $i F = F i : F F \to F$.

A subfunctor of the identity $F$ is called idempotent if $i F = F i: F F\to F$ is an isomorphism.

In Abelian context, consider additive subfunctors $\sigma$ of the identity.

If $\sigma$ is left exact, then it is idempotent.

For the proof, start with the exact sequence

$$0\to \sigma(M)\overset{i_M}\hookrightarrow M\to M/\sigma(M)\overset{p}\to 0$$

By left exactness of $\sigma$,

$$0\to \sigma(\sigma(M))\overset{i_{\sigma(M)}}\hookrightarrow \sigma(M)\overset{\sigma(p)}\to \sigma(M/\sigma(M))\to 0$$

Now we decompose $\sigma(p)$

$$\begin{array}{ccc} \sigma(M) &\overset{i_M}\hookrightarrow & M \\ \sigma(p)\downarrow & & \downarrow p \\ \sigma(M/\sigma(M)) & = & M/\sigma(M), \end{array}$$

hence $\sigma(p) = p\circ i_M = 0$ and therefore $\sigma(\sigma(M)) = \sigma(M)$.

Left exactness implies more generally that for any $M\subset N$, $\sigma(M) = M\cap \sigma(N)$.

Idempotent kernel functor (for some: radical functor) is left exact subfunctor of the identity satisfying $\sigma(M/\sigma(M)) = 0$. Identity $\sigma(M/\sigma(M)) = 0$ holds for Jacobson radical, which is in general neither idempotent nor left exact.

Let $L\subset \sigma(M)$. The commutative diagram

$$\begin{array}{ccc} \sigma(M) &\overset{i_M}\hookrightarrow & M \\ \sigma(q) \downarrow & & \downarrow q \\ \sigma(M/L) & \overset{i_{M/L}}\hookrightarrow & M/L, \end{array}$$

shows $Ker(\sigma(q)) = Ker(q\circ i_M) = Ker(q) = L$; therefore $\sigma(M/L)\subset \sigma(M)/L$. On the other hand,

$$\begin{array}{ccc} \sigma(M/L) &\overset{i_{M/L}}\hookrightarrow & M/L \\ \sigma(p') \downarrow & & \downarrow p' \\ \sigma(M/\sigma(M)) & \overset{i_{M/L}}\hookrightarrow & M/\sigma(M), \end{array}$$

and if $\sigma(M/\sigma(M))=0$ then $p'|_{\sigma(M/L)} = 0$, that is $\sigma(M/L)\subset Ker(M/L\to M/\sigma(M)) = \sigma(M)/L$.

Summarizing, if $\sigma(M/\sigma(M)) = 0$ then $\sigma(M/L) = \sigma(M)/L$.