# In a category with epi-equalizer factorization , composition of equalizers is an equalizer

The problem says : (in a category with epi-equalizer factorizations)

if $gf$ is an equalizer show that $f$ is an equalizer , using this result, also show that if $f$ and $g$ are equalizers then $gf$ is an equalizer.

What I've found so far:

• In this question there is a counterexample in a category without (epi-equalizer) factorizations.

My attempt of a solution ($gf$ equalizer $\rightarrow$ $f$ equalizer ):

Since $gf$ is an equalizer it has to be the equalizer for some arrows $\alpha$ ,$\beta$ , I hope to show that $f$ is the equalizer for $\alpha g$ , $\beta g$

We have $\alpha(gf) = \beta(gf)$ , then $(\alpha g) f = (\beta g) f$.

Let $\phi$ be another candidate for equalizer so $(\alpha g) \phi = (\beta g) \phi$. !

Since $gf$ is an equalizer and $(\alpha g) \phi = (\beta g) \phi$ , there exists an unique arrow $\psi$ which satisfies $gf\psi = g\phi$ !

Let $e,i$ be a (epi-equalizer) factorization of $g$ .

We have $ief\psi = ie\phi$ , since $i$ is an equalizer it's also mono, so $ef\psi = e\phi$.

The picture would look like this : !

This is where I can't continue , if I could make that $e$ go away from the last equation I'd get the existence of the arrow ($f\psi = \phi$) , the uniqueness would follow from factoring f. I have tried to factor $\phi$ or $gf$ but I couldn't see how it would help me.

So far , I haven't used that $e$ is epi or that $i$ is an equalizer (only that it's mono)

• I know a proof assuming that regular monomorphisms are closed under pushout. I don't know if the claim is true otherwise. – Zhen Lin Sep 27 '15 at 8:24

## 2 Answers

First as clarification: If $f$ is a morphism we say $id_{d(f)}$ is the identity morphism on the domain of $f$ and $id_{c(f)}$ is the identity on the codomain of $f$.

lemma: Let be $C$ a category with epic-equalizer factorization. Let $f$ be a morphism and $e_f$, $i_f$ its epic-equalizer factorization.

$f$ is equalizer if and only if $e_f$ is an isomorphism.

Proof: If $f$ is equalizer, $id_{d(f)} f = f$ is a epic-equalizer factorization. For the universal property there is an unique isomorphism $\phi$ such that $e_f = \phi id_{d(f)} = \phi$ therefore $e_f$ is an isomorphism.

If $e_f$ is an isomorphism. Let $a$ and $b$ the morphisms $i_f$ equalizes. We claim $f$ is an equalizer of $a$ and $b$. $$a i_f = bi_f$$ $$a i_f e_f = b i_f e_f$$ $$a f = bf$$ Let $k$ be a morphism such that $ak = bk$ for the universal property of $i_f$ there is a unique $\psi$ such that $i_f \psi = k$ or $k = i_f id_{c(e_f)} \psi = i_f e_f(e_f^{-1} \psi ) = f \psi^{\prime}$. Let's see $\psi^{\prime} = e_f^{-1} \psi$ is unique.

If there is a $\phi$ such that $f \phi = k$ $$i_f e_f \phi = i_fe_f e_f^{-1} \psi$$ $$e_f \phi = e_f e_f^{-1} \psi$$ $$\phi = e_f^{-1} \psi = \psi^{\prime}$$ $f$ is equalizer and we are done.

Now the main result: Let $i_g e_g$ and $i_f e_f$ the factorizations epic-equalizer of $g$ and $f$ respectively.

$$\alpha gf = \beta gf$$ $$\alpha g i_f e_f = \beta g i_f e_f$$ $$\alpha g i_f = \beta g i_f$$ for the universal property of $gf$ as equalizer there is a unique $\psi$ such that $gf \psi = g i_f$ if we add $e_f$, $gf (\psi e_f) = gf id_{d(f)}$. Remeber that $gf$ is an equalizer therefore a monomorphism what it means: $\psi e_f = id_{d(f)}= id_{d(e_f)}$ or, what is the same, ¡$e_f$ is a split- monomorphism then equalizer! and we know that if a morphism is equalizer and epic it is isomorphism. Therefore we are done.

Factorise $f$ instead: say $f = i \circ e$ with $i: I \to Y$ an equaliser and $e: X \to I$ epic.

# Claim: $i$ is the equaliser of $\alpha g, \beta g$.

Indeed, let $z: Z \to Y$ be such that $\alpha g z = \beta g z$. We want to find unique $\bar{z}: z \to I$ such that $i \bar{z} = z$.

Have $g z$ equalises $\alpha, \beta$ so it lifts to unique $\overline{gz}: Z \to C$ such that $gf \circ \overline{gz} = gz$: so we have found a candidate $\bar{z}: Z \to I$ given by $\bar{z} = e \circ \overline{gz}$. It is in fact unique: let $k: Z \to I$ also have $ik = z = i e \overline{g z}$. $i$ is an equaliser and hence monic, so $k = e \overline{g z}$ as required.

Therefore $f = i e$ where $i$ is the equaliser of $\alpha g, \beta g$.

# Claim: $f$ is (also) the equaliser of $\alpha g, \beta g$.

Indeed, let $r: R \to Z$ be any arrow such that $\alpha r = \beta r$. Then $r$ lifts uniquely to $\bar{r}: R \to X$ such that $gf\bar{r} = r$. Then $f\bar{r}: R \to Y$ has the same effect on $\alpha g$ as $\beta g$, so $f \bar{r}$ lifts to a unique $r': R \to I$ with $i r' = f \bar{r}$. Since $f = i e$ and $i$ is an equaliser (and hence monic), this is unique $r': R \to I$ with $r' = e \bar{r}$. This is summarised in the diagram, where the two dotted arrows are unique such that the entire diagram commutes. Apologies for the awful quality. But now we see that $f$ must also equalise $\alpha g, \beta g$, because any arrow into $Y$ setting $\alpha g = \beta g$ lifts to $I$ and thence to $X$.

I think the foregoing reasoning is sound. Do check it thoroughly yourself.

• Hello @Patrick, thanks for you answer. There's only one step in the middle I don't understand, when you found the candidate $\overline{z} = e \overline{gz}$ , I agree that $gf \overline{gz} = gz$ which gives us (factoring f) $gie\overline{gz} = gz$.but we needed to show that $ie\overline{gz} = z$. which you use on the very next line, ¿why is the last equation true?. – sr chunchurria Sep 26 '15 at 23:30
• Oh dear. I've just woken up and I have no idea whether I justified this step to myself yesterday. I'll think about it, but now I'm worried it isn't justified. – Patrick Stevens Sep 27 '15 at 7:41