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Suppose $gf$ is an equalizer in a category $\mathfrak C$, I think that $f$ not necessarly is an equalizer, but I don't know how to come up with a counterexample; i've really tried it so hard. Thanks for any help.

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2 Answers 2

up vote 5 down vote accepted

(Third time's the charm? -- old wrong answer left deleted because its comments don't apply here)

Consider the following category with objects $\{1,2,3,4,5\}$.

1 --s--> 2 --f--> 3 --g--> 4 --p--> 5
  -------h------>            --q-->

There's an arrow $n\to m$ whenever $n\le m$, and these arrows are unique except for the following cases:

  • $fs \ne h: 1\to 3$
  • $p \ne q: 4\to5$
  • $pg \ne qg : 3 \to 5$

Then $gf$ is an equalizer of $p$ and $q$, as seen by inspecting all arrows that end at $4$:

  • $g$ and $\mathrm{id}_4$ are out because $pg\ne qg$.
  • $gf$ is the equalizer itself.
  • $gfs$ satisfies $p(gfs)=q(gfs)$. It factors through $gf$ as it should, and the mediating arrow $s$ is trivially unique.
  • $gh$ is the same arrow as $gfs$.

However, $f$ is not an equalizer. In particular it is not an equalizer of $pg$ and $qg$ because $(pg)h=(qg)h$ yet $h$ does not factor through $f$.

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1+. Thanks for providing the answer which my comment refers to ;). One also has to check that your category is, indeed, a category. This is not completely trivial, because essentially it means to find a normal form of a certain category defined by generators and relations. –  Martin Brandenburg Feb 12 '13 at 21:57
1  
@MartinBrandenburg: Is there anything to check but associativity? That is trivial here because every $a\circ b\circ c$ either involves an identity or is so long that its result belongs to a hom-set that I've defined to be a singleton. –  Henning Makholm Feb 12 '13 at 22:01
    
Yes, you're right. –  Martin Brandenburg Feb 13 '13 at 0:47

In many cases this property of regular monomorphisms actually holds.

  1. A trivial but already quite large class of examples is where every monomorphism is regular.

  2. If $gf$ is a regular monomorphism and $g$ is a monomorphism, then $f$ is a regular monomorphism. Namely, if $gf$ is the equalizer of $u,v$, then one can check that $f$ is the equalizer of $ug,vg$.

  3. If $gf$ is a strong monomorphism, then also $f$ is a strong monomorphism. This is an easy observation, a proof can be found in Borceux, Handbook of categorical algebra, Volume 1, Proposition 4.6.5 (2).

  4. Every regular epimorphism is strong (loc. cit., part (4) of that Proposition). The converse holds in every regular category: Borceux, Handbook of categorical algebra, Volume 2, Proposition 2.1.4.

Hence, in a coregular category (= dual to a regular category), regular monomorphisms coincide with strong monomorphisms, which implies the desired property. Besides from existence of certain (co)limits, a category is coregular when regular monomorphisms are universal, i.e. if $A \hookrightarrow B$ is a regular monomorphism and $A \to C$ is an arbitrary morphism ("cobase change"), then also $C \to B \cup_A C$ is a regular monomorphism. There are lots of examples, for example $\mathsf{Set}$, $\mathsf{Top}$, $\mathsf{Haus}$ abelian categories, and $\mathsf{Ban}_1$. Is $\mathsf{Grp}$ is coregular?

What makes me wonder is that Proposition 6.4 in the paper Adhesive and quasiadhesive categories by Stephen Lack and Pawel Sobicinski reads: "The following hold in any category C: (i) if mn is a regular monomorphism and m is arbitrary then n is a regular monomorphism;" I think that Henning's counterexample shows that this is wrong.


Addendum. The following lemma often appears in the foundations of algebraic geometry, but it is also useful here.

Lemma. Let $P$ be a class of morphisms in a category with pullbacks which is stable under pullbacks and composition. Also assume that every diagonal morphism $Y \to Y \times_S Y$ lies in $P$. Then $gf \in P$ implies $f \in P$ (cancellation property).

Proof. Write $X \xrightarrow{f} Y \xrightarrow{g} S$ and factor $f$ as $X \xrightarrow{\Gamma_f} X \times_S Y \xrightarrow{\mathrm{pr}_2} Y$. Here, $\Gamma_f$ is a pullback of the diagonal $Y \to Y \times_S Y$, and $\mathrm{pr}_1$ is a pullback of $X \to S$. Both are in $P$. $\square$

In a category, regular monomorphisms are always stable under pullbacks (Handbook, Vol. 1, Prop. 4.3.8 (2)), and split monomorphisms are always regular. This proves:

Corollary. In a category with pullbacks such that regular monomorphisms are closed under composition, the regular monomorphisms satisfy the cancellation property.

This applies to a large class of categories, among of which are coregular categories, but many more.

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Adamek-Herrlich-Strecker, Abstract and Concrete Categories, have some examples. In the exercises to section 7 they ask the reader to verify examples showing that: (7J) in the category of functionally Hausdorff spaces, regular monomorphisms are not closed under composition and $gf$ regular monic does not imply $f$ regular monic; dually (7S) in $\mathsf{Cat}$ regular epis are not closed under composition and $gf$ regular epi does not imply $g$ regular epi. (I didn't check the details myself) –  Martin Feb 13 '13 at 3:40
    
Why not adding this as an answer? This is not a comment to my answer, it is an answer. –  Martin Brandenburg Feb 13 '13 at 13:50

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