Zoran Skoda subfunctor of identity

Let AA be a concrete category.

An endofunctor F:AAF : A\to A is a subfunctor of the identity functor if F(M)MF(M)\subset M for each object MM in AA. That means in particular that the inclusion is a natural transformation. Thus, if g:MNg:M\to N is a morphism in AA, F(g)F(g) is simply the restriction-corestriction of gg and g(F(M))=F(g)(M)F(N)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)(F,i) where FF is an endofunctor and i:FId Ai:F\Rightarrow Id_A is the natural transformations whose components i M:F(M)Mi_M:F(M)\hookrightarrow M are monic. The commutative diagram

F(M) i M M F(g) g F(N) i N N \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 gi M:F(M)Ng\circ i_M:F(M)\to N of GG and then, by monicity of i Ni_N, determines the arrow F(M)F(N)F(M)\to F(N) uniquely; this is the restriction-corestriction when it exists.

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

If g=i Mg = i_M then by monicity of i Mi_M from the following commutative diagram

FFM i FM FM F(i M) i M FM i M M \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 FMF(i_M) = (i_M)|_{F(M)} = i_{F M}, that is, iF=Fi:FFFi F = F i : F F \to F.

A subfunctor of the identity FF is called idempotent if iF=Fi:FFFi F = F i: F F\to F is an isomorphism.

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

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

For the proof, start with the exact sequence

0σ(M)i MMM/σ(M)p0 0\to \sigma(M)\overset{i_M}\hookrightarrow M\to M/\sigma(M)\overset{p}\to 0

By left exactness of σ\sigma,

0σ(σ(M))i σ(M)σ(M)σ(p)σ(M/σ(M))0 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 σ(p)\sigma(p)

σ(M) i M M σ(p) p σ(M/σ(M)) = M/σ(M),\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 σ(p)=pi M=0\sigma(p) = p\circ i_M = 0 and therefore σ(σ(M))=σ(M)\sigma(\sigma(M)) = \sigma(M).

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

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

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

σ(M) i M M σ(q) q σ(M/L) i M/L M/L,\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(σ(q))=Ker(qi M)=Ker(q)=LKer(\sigma(q)) = Ker(q\circ i_M) = Ker(q) = L; therefore σ(M/L)σ(M)/L\sigma(M/L)\subset \sigma(M)/L. On the other hand,

σ(M/L) i M/L M/L σ(p) p σ(M/σ(M)) i M/L M/σ(M),\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 σ(M/σ(M))=0\sigma(M/\sigma(M))=0 then p| σ(M/L)=0p'|_{\sigma(M/L)} = 0, that is σ(M/L)Ker(M/LM/σ(M))=σ(M)/L\sigma(M/L)\subset Ker(M/L\to M/\sigma(M)) = \sigma(M)/L.

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

Last revised on August 20, 2024 at 21:53:01. See the history of this page for a list of all contributions to it.