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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 $(\alpha g) \phi = (\beta g) \phi$.

a busy cat!

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

a busy cat!

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 :

a busy cat!

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)

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

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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.

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.

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  • $\begingroup$ 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?. $\endgroup$ – sr chunchurria Sep 26 '15 at 23:30
  • $\begingroup$ 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. $\endgroup$ – Patrick Stevens Sep 27 '15 at 7:41

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