regular monomorphism in nLab

Context

Category theory

Higher category theory

• Idea
• Definition
• Properties
• Relation to effective monomorphisms
• Examples
• In an (∞,1)(\infty,1)-category
• Related pages
• References

Definition

A monomorphism i:A→Bi: A \to B is an effective monomorphism if it is the equalizer of its cokernel pair: if the pushout

A →i B i↓ ↓i 1 B →i 2 B+ AB\array{ A & \stackrel{i}{\to} & B \\ i \downarrow & & \downarrow i_1 \\ B & \underset{i_2}{\to} & B +_A B }

exists and ii is the equalizer of the pair of coprojections i 1,i 2:B→→B+ ABi_1, i_2: B \stackrel{\to}{\to} B +_A B. Obviously effective monomorphisms are regular.

Proof

It is clear that every effective monomorphism is regular, we need to show the converse.

Suppose i:A→Bi \colon A \to B is the equalizer of a pair of morphisms f,g:B→Cf, g: B \to C, and with notation as in def. , let j:E→Bj: E \to B be the equalizer of the pair of coprojections i 1,i 2i_1, i_2. Since f∘i=g∘if \circ i = g \circ i, there exists a unique map ϕ:B+ AB→C\phi: B +_A B \to C such that ϕ∘i 1=f\phi \circ i_1 = f and ϕ∘i 2=g\phi \circ i_2 = g. Then, since

fj=ϕi 1j=ϕi 2j=gjf j = \phi i_1 j = \phi i_2 j = g j

and since i:A→Bi: A \to B is the equalizer of the pair (f,g)(f, g), there is a unique map k:E→Ak: E \to A such that j=ikj = i k. Since i 1i=i 2ii_1 i = i_2 i, there is a unique map l:A→El: A \to E such that i=jli = j l. The maps kk, ll are mutually inverse.

Proof

Regarding the first statement: an injective continuous function f:X→Yf \colon X \to Y clearly has the cancellation property that defines monomorphisms: for parallel continuous functions g 1,g 2:Z→Xg_1,g_2 \colon Z \to X: if f∘g 1=f∘g 1f \circ g_1 = f \circ g_1, then g 1=g 2g_1 = g_2 because continuous functions are equal precisely if their underlying functions of sets are equal. Conversely, if ff has the cancellation property, then testing on points g 1,g 2:*→Xg_1, g_2 \colon \ast \to X gives that ff is injective.

Regarding the second statement: from the construction of equalizers in Top (this example) we have that these are topological subspace inclusions.

Conversely, let i:X→Yi \colon X \to Y be a topological subspace embedding. We need to show that this is the equalizer of some pair of parallel morphisms.

To that end, form the cokernel pair (i 1,i 2)(i_1, i_2) by taking the pushout of ii against itself (in the category of sets, and using the quotient topology on a disjoint union space). By prop. , the equalizer of that pair is the set-theoretic equalizer of that pair of functions endowed with the subspace topology. Since monomorphisms in Set are regular, we get the function ii back and (again by this example) it is equipped with the subspace topology.

Proposition

In Grp, the monics are (up to isomorphism) the inclusions of subgroups, and every monomorphism is regular.

Proof

The elementary proof we give follows exercise 7H of (AdamekHerrlichStrecker). It is however nonconstructive (because it contains if-then-else lines); for a constructive proof, see here.

Let K↪HK \hookrightarrow H be a subgroup. We need to define another group GG and group homomorphisms f 1,f 2:H→Gf_1, f_2 : H \to G such that

K={h∈H|f 1(h)=f 2(h)}. K = \{h \in H | f_1(h) = f_2(h)\} \,.

To that end, let

X:=H/K∐{K^}:={hK|h∈H}∐{K^} X := H/K \coprod \{\hat K\} := \{ h K | h \in H \} \coprod \{\hat K\}

be the set of cosets together with one more element K^\hat K.

Let then

G=Aut Set(X) G = Aut_{Set}(X)

be the permutation group on XX.

Define ρ∈G\rho \in G to be the permutation that exchanges the coset eKe K with the extra element K^\hat K and is the identity on all other elements.

Finally define group homomorphism f 1,f 2:H→Gf_1,f_2 : H \to G by

f 1(h):x↦{hh′K ifx=h′K K^ ifx=K^ f_1(h) : x \mapsto \left\{ \array{ h h' K & if x = h' K \\ \hat K & if x = \hat K } \right.

and

f 2(h)=ρ∘f 1(h)∘ρ −1. f_2(h) = \rho \circ f_1(h) \circ \rho^{-1} \,.

It is clear that these maps are indeed group homomorphisms.

So for h∈Hh \in H we have that

f 1(h):K^↦K^, f_1(h) : \hat K \mapsto \hat K \,,

and

f 1(h):eK↦hK f_1(h) : e K \mapsto h K

and

f 2(h):K^↦eK↦hK↦{K^ ifh∈K hK otherwise. f_2(h) : \hat K \mapsto e K \mapsto h K \mapsto \left\{ \array{ \hat K & if h \in K \\ h K & otherwise } \right. \,.

f 2(h):eK↦K^↦K^↦eK. f_2(h) : e K \mapsto \hat K \mapsto \hat K \mapsto e K \,.

So we have f 1(h)=f 2(h)f_1(h) = f_2(h) precisely if h∈Kh \in K.