A monomorphism is regular if it behaves like an embedding.
effective epimorphism$\Rightarrow$ regular epimorphism $\Leftrightarrow$ covering
effective monomorphism$\Rightarrow$ regular monomorphism $\Leftrightarrow$ embedding .
The universal factorization through an embedding is the image.
A regular monomorphism is a morphism $f : c \to d$ in some category which occurs as the equalizer of some pair of parallel morphisms $d \stackrel{\to}{\to} e$, i.e. for which a limit diagram of the form
exists.
From the defining universal property of the limit it follows directly that a regular monomorphism is in particular a monomorphism.
The dual concept is that of a regular epimorphism.
Beware that (CassidyHebertKelly) use ‘regular monomorphism’ in a more general way: for them, a regular monomorphism is by definition the joint equalizer of an arbitrary family of parallel pairs of morphisms with common domain. This concept is sometimes called strict monomorphism, dual to the more commonly used strict epimorphism.
A monomorphism $i: A \to B$ is an effective monomorphism if it is the equalizer of its cokernel pair: if the pushout
exists and $i$ is the equalizer of the pair of coprojections $i_1, i_2: B \stackrel{\to}{\to} B +_A B$. Obviously effective monomorphisms are regular.
In a category with equalizers and cokernel pairs, the class of regular monomorphism coincides with that of effective monomorphism (def. ).
It is clear that every effective monomorphism is regular, we need to show the converse.
Suppose $i \colon A \to B$ is the equalizer of a pair of morphisms $f, g: B \to C$, and with notation as in def. , let $j: E \to B$ be the equalizer of the pair of coprojections $i_1, i_2$. Since $f \circ i = g \circ i$, there exists a unique map $\phi: B +_A B \to C$ such that $\phi \circ i_1 = f$ and $\phi \circ i_2 = g$. Then, since
and since $i: A \to B$ is the equalizer of the pair $(f, g)$, there is a unique map $k: E \to A$ such that $j = i k$. Since $i_1 i = i_2 i$, there is a unique map $l: A \to E$ such that $i = j l$. The maps $k$, $l$ are mutually inverse.
In Set, or more generally in any pretopos, every monomorphism is regular.
Similarly, in Ab, and more generally any abelian category, every monomorphism is regular.
(regular monomorphisms of topological spaces)
In the category Top of topological space,
the monomorphisms are the those continuous functions which are injective functions;
the regular monomorphisms are the topological embeddings (that is, the injective continuous functions whose sources have the topologies induced from their targets); these are in fact all of the extremal monomorphisms.
Regarding the first statement: an injective continuous function $f \colon X \to Y$ clearly has the cancellation property that defines monomorphisms: for parallel continuous functions $g_1,g_2 \colon Z \to X$: if $f \circ g_1 = f \circ g_1$, then $g_1 = g_2$ because continuous functions are equal precisely if their underlying functions of sets are equal. Conversely, if $f$ has the cancellation property, then testing on points $g_1, g_2 \colon \ast \to X$ gives that $f$ 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 $i \colon X \to Y$ 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 $(i_1, i_2)$ by taking the pushout of $i$ 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 $i$ back and (again by this example) it is equipped with the subspace topology.
In Grp, the monics are (up to isomorphism) the inclusions of subgroups, and every monomorphism is regular.
In contrast, the normal monomorphisms (where one of the morphisms $d \to e$ is required to be the zero morphism) are the inclusions of normal subgroups.
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 $K \hookrightarrow H$ be a subgroup. We need to define another group $G$ and group homomorphisms $f_1, f_2 : H \to G$ such that
To that end, let
be the set of cosets together with one more element $\hat K$.
Let then
be the permutation group on $X$.
Define $\rho \in G$ to be the permutation that exchanges the coset $e K$ with the extra element $\hat K$ and is the identity on all other elements.
Finally define group homomorphism $f_1,f_2 : H \to G$ by
and
It is clear that these maps are indeed group homomorphisms.
So for $h \in H$ we have that
and
and
So we have $f_1(h) = f_2(h)$ precisely if $h \in K$.
In the context of higher category theory the ordinary limit diagram $c \stackrel{f}{\to} d \stackrel{\to}{\to} e$ may be thought of as the beginning of a homotopy limit diagram over a cosimplicial diagram
Accordingly, it is not unreasonable to define a regular monomorphism in an (∞,1)-category, to be a morphism which is the limit in a quasi-category of a cosimplicial diagram.
In practice this is of particular relevance for the $\infty$-version of regular epimorphisms: with the analogous definition as described there, a morphism $f : c \to d$ is a regular epimorphism in an (∞,1)-category $C$ if for all objects $e \in C$ the induced morphism $f^* : C(d,e) \to C(c,e)$ is a regular monomorphism in ∞Grpd (for instance modeled by a homotopy limit over a cosimplicial diagram in SSet).
Warning. The same warning as at regular epimorphism applies: with this definition of regular monomorphism in an (∞,1)-category these may fail to satisfy various definitions of plain monomorphisms that one might think of.
Textbook accounts:
See also:
Last revised on May 14, 2022 at 12:26:11. See the history of this page for a list of all contributions to it.