If $gf$ is an equalizer , is $f$ an equalizer? 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. 
 A: (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 $f$.
A: In many cases this property of regular monomorphisms actually holds.


*

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

*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 $ug,vg$.

*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).

*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.
