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Stream: theory: mathematics

Topic: Is this paper serious?

Jean-Baptiste Vienney (Jan 26 2025 at 19:44):

In this paper: Coverings and ring-groupoids, it is said that the fundamental groupoid of a topological ring is a ring-groupoid i.e. a ring-object in the category of groupoids. But the fundamental groupoid of a topological ring is a rather trivial object since it is a thin category.

Indeed: Let $R$ be a topological ring. First, we show that $\pi^1(R,0) \simeq \{1\}$.

Let $\gamma:[0,1] \rightarrow R$ be a path from $0$ to $r \in R$. Define $H:[0,1] \times [0,1] \rightarrow R$ by $H(s,t)=i(t)\gamma(s)$. Then $H$ is a continuous map. Moreover, $H(s,0)=i(0)\gamma(s)=0\gamma(s)=0$ and $H(s,1)=i(1)\gamma(s)=1\gamma(s)=\gamma(s)$. We deduce that $[\gamma]=[e_0]$.

Now, let $a,b \in R$, we have an homeomorphism $u:R \rightarrow R$ given by $u(r)=r-a$. It gives a group isomorphism $\phi:\pi_1(R,a) \simeq \pi_1(R,0)$ defined by $\phi([\gamma])=[\gamma-a]$ with inverse given by $\phi^{-1}([\tau])=[\tau+a]$. We deduce that $\pi_1(R,a) \simeq \{1\}$ for every $a \in R$.

Finally, let $a,b \in R$ and let $\gamma,\gamma':[0,1] \rightarrow R$ be two paths from $a$ to $b$. We obtain a path $\gamma * (\gamma')^{-1}$ from $a$ to $a$. We have $[\gamma * (\gamma')^{-1}]=[\gamma][\gamma']^{-1}=1$. Thus $[\gamma]=[\gamma']$.

We conclude that the fundamental groupoid of $R$ is isomorphic as a category to the poset whose objects are path-components of $R$ and where $a \sim b$ iff $a$ and $b$ are in the same path-component.

Should I conclude that this paper does not contain any interesting content? Maybe there is something interesting that I don't understand or maybe my proof above is wrong.

Jean-Baptiste Vienney (Jan 26 2025 at 19:46):

I'm even more intrigued when I see that this paper has been cited 33 times.

John Baez (Jan 26 2025 at 19:50):

You're reminding me of this:

• Tom Leinster, Holy crap, do you know what a compact ring is?

This result implies that a compact Hausdorff topological ring has a fundamental groupoid whose only morphisms are identity morphisms.

Jean-Baptiste Vienney (Jan 26 2025 at 19:55):

Thanks, this post looks interesting!

John Baez (Jan 26 2025 at 19:55):

Jean-Baptiste Vienney said:

But the fundamental groupoid of a topological is a rather trivial object since it is a thin category.

You left out the word "ring".

Indeed: Let $R$ be a topological ring. First, we show that $\pi^1(R,0) \simeq \{1\}$.

Let $\gamma:[0,1] \rightarrow R$ be a path from $0$ to $r \in R$. Define $H:[0,1] \times [0,1] \rightarrow R$ by $H(s,t)=i(t)\gamma(s)$.

What's $i$?

Jean-Baptiste Vienney (Jan 26 2025 at 19:58):

Argh, I first proved that the fundamental group of a path-connected topological ring is trivial where I used a path $i$ from $0$ to $1$ in $R$. And now I'm still using this path $i$ so that my proof is not correct!!

Jean-Baptiste Vienney (Jan 26 2025 at 19:59):

Well-spoted.

Jean-Baptiste Vienney (Jan 26 2025 at 20:00):

So I'm back to being confused about whether the fundamental groups and groupoids of topological rings are interesting or not.

Jean-Baptiste Vienney (Jan 26 2025 at 20:04):

I think I'm still right that the fundamental group of a path-connected topological ring is trivial.

Jean-Baptiste Vienney (Jan 26 2025 at 20:06):

However the fundamental group of a path-component of a topological ring should not necessarily be trivial.

John Baez (Jan 26 2025 at 20:17):

Jean-Baptiste Vienney said:

So I'm back to being confused about whether the fundamental groups and groupoids of topological rings are interesting or not.

That's a fascinating question!

I can't think of any topological rings with fundamental groupoids that are not equivalence relations (=thin preorders). Tom showed they must be equivalence relations for topological rings that are compact Hausdorff. Have you proved it for topological rings that have a path from 0 to 1?

Jean-Baptiste Vienney (Jan 26 2025 at 20:21):

Yes, let me write the proof here again.

Jean-Baptiste Vienney (Jan 26 2025 at 20:22):

Wait, I must think about what is the proposition too.

John Baez (Jan 26 2025 at 20:23):

Sure, don't rush.

Jean-Baptiste Vienney (Jan 26 2025 at 20:23):

Proposition: Let $R$ be a topological ring such that $0$ and $1$ are in the same path component. Then $\pi_1(R,0)$ is trivial.

Jean-Baptiste Vienney (Jan 26 2025 at 20:27):

Proof: Let $i:[0,1] \rightarrow R$ be a path from $0$ to $1$ in $R$. Let $x \in R$ and let $\gamma:[0,1] \rightarrow R$ be a path from $0$ to $x$. Define $H:[0,1] \times [0,1] \rightarrow R$ by $H(s,t)=i(t)\gamma(s)$. Then, $H$ is a continuous map. Moreover, $H(s,0)=i(0)\gamma(s)=0$, $H(s,1)=i(1)\gamma(s)=\gamma(s)$ so that $[\gamma]=1$ the identity of $\pi_1(R,0)$.

Jean-Baptiste Vienney (Jan 26 2025 at 20:28):

$\pi_1(R,0)$ is the fundamental group of $R$ based at $0 \in R$.

John Baez (Jan 26 2025 at 20:29):

Great! So this implies that $\pi_1(R,r)$ is trivial for any basepoint $r \in R$.

Jean-Baptiste Vienney (Jan 26 2025 at 20:30):

Yes, I agree.

John Baez (Jan 26 2025 at 20:30):

(You can use addition to translate a homotopy class of loops from $r$ to $0$.)

John Baez (Jan 26 2025 at 20:31):

And that implies that the fundamental groupoid of a a topological ring with a path from $0$ to $1$ must be equivalent to a discrete category (one with only identity morphisms).

Jean-Baptiste Vienney (Jan 26 2025 at 20:31):

Yes, exactly.

John Baez (Jan 26 2025 at 20:32):

I also think this result implies that any topological field has a fundamental groupoid that's equivalent to a discrete category, since in a topological field any continuous path from $x$ to $y$ with $y \ne x$ can be turned into a continuous path from $0$ to $1$.

John Baez (Jan 26 2025 at 20:33):

So I am fairly optimistic that all topological rings have a fundamental groupoid that's equivalent to a discrete groupoid.... and if there's a counterexample I'd really like to see it!

Jean-Baptiste Vienney (Jan 26 2025 at 20:34):

John Baez said:

And that implies that the fundamental groupoid of a a topological ring with a path from $0$ to $1$ must be equivalent to a discrete category (one with only identity morphisms).

Did you use that every thin category is equivalent to a discrete category? Is it really true? Because what I proved at the beginning finally is that if a topological ring has a path from $0$ to $1$ then its fundamental groupoid is a thin category.

Jean-Baptiste Vienney (Jan 26 2025 at 20:36):

Oh but this fundamental groupoid is a thin groupoid not just a thin category.

Jean-Baptiste Vienney (Jan 26 2025 at 20:36):

And every thin groupoid is equivalent to a discrete category?

John Baez (Jan 26 2025 at 20:36):

Did you use that every thin category is equivalent to a discrete category? Is it really true?

No, that's not true: any preorder is a thin category.

But the fundamental groupoid is a groupoid, and any thin groupoid is equivalent to a discrete category.

Jean-Baptiste Vienney (Jan 26 2025 at 20:39):

John Baez said:

I also think this result implies that any topological field has a fundamental groupoid that's equivalent to a discrete category, since in a topological field any continuous path from $x$ to $y$ with $y \ne x$ can be turned into a continuous path from $0$ to $1$.

Wait, I think that more precisely if a topological field has a nontrivial path component, then its fundamental groupoid is equivalent to a discrete category.

John Baez (Jan 26 2025 at 20:40):

A thin groupoid is a fancy name for a category where there's at most one morphism from any object to any other object, and this morphism is invertible. So it's a category where any two objects either have no morphisms between them, or exactly one isomorphism. So it's equivalent to the discrete category on its set of isomorphism classes.

Jean-Baptiste Vienney (Jan 26 2025 at 20:40):

John Baez said:

A thin groupoid is a fancy name for a category where there's at most one morphism from any object to any other object, and this morphism is invertible. So it's a category where any two objects either have no morphisms between them, or exactly one isomorphism. So it's equivalent to the discrete category on its set of isomorphism classes.

Ok, thanks.

John Baez (Jan 26 2025 at 20:40):

Jean-Baptiste Vienney said:

John Baez said:

I also think this result implies that any topological field has a fundamental groupoid that's equivalent to a discrete category, since in a topological field any continuous path from $x$ to $y$ with $y \ne x$ can be turned into a continuous path from $0$ to $1$.

Wait, I think that more precisely if a topological field has a nontrivial path component, then its fundamental groupoid is equivalent to a discrete category.

That's half of the argument.

Jean-Baptiste Vienney (Jan 26 2025 at 20:41):

What is the second half?

John Baez (Jan 26 2025 at 20:42):

I'm saying that in a topological field there's either no continuous path from any element $x$ to any $y \ne x$ (in which case the fundamental groupoid is equivalent to a discrete category) or there is such a path (in which case there's a path from $0$ to $1$, so your argument shows the fundamental groupoid is equivalent to a discrete category).

Jean-Baptiste Vienney (Jan 26 2025 at 20:44):

Ok, that's fine.

Jean-Baptiste Vienney (Jan 26 2025 at 20:46):

I feel it makes more natural the fact that a lot of topological fields are totally disconnected.

Jean-Baptiste Vienney (Jan 26 2025 at 20:48):

(At least, it seems related.)

Jean-Baptiste Vienney (Jan 26 2025 at 20:59):

So the fundamental groupoid of a topological field is always equivalent to a discrete category. I've just understood your reasoning :sweat_smile: .

Jean-Baptiste Vienney (Jan 26 2025 at 21:02):

If there is no path from any $x$ to a $y \neq x$ then every path from $x$ to $x$ is equal to the constant path, else it would give a path from $x$ to some $y \neq x$.

Jean-Baptiste Vienney (Jan 26 2025 at 21:05):

It must tell us something about the fundamental groupoid of a topological integral domain.

John Baez (Jan 26 2025 at 21:05):

Great! No worries.

By the way, it's possible that you could continue your argument to show that if a topological ring $R$ has a (continuous) path from $0$ to $1$ then all its homotopy groups $\pi_n(R,r)$ are trivial. In this case we say $R$ is "weakly homotopy equivalent" to a discrete topological space.

John Baez (Jan 26 2025 at 21:08):

I think it would be fun to prove that every topological field is weakly homotopy equivalent to a discrete space.

And I almost feel like conjecturing that any topological ring is weakly homotopy equivalent to a discrete space! Certainly Tom showed any compact Hausdorff topological ring has this property - because they are totally disconnected. We have topological rings like $\mathbb{R}$ and $\mathbb{C}$ and $\mathbb{R}[x]$ and $\mathbb{R}[x,y]/\langle x^2 - 3y \rangle$ that are not totally disconnected, but they're contractible, so they're weakly homotopy equivalent to a point.

Jean-Baptiste Vienney (Jan 26 2025 at 21:15):

I've never worked with higher homotopy groups ahah. You can think about it yourself if you want. Just say that I've contributed :sweat_smile:

Jean-Baptiste Vienney (Jan 26 2025 at 21:17):

I don't know how to handle a continuous maps $S^n \rightarrow R$. Say I have a continuous path $i:[0,1] \rightarrow R$ from $0$ to $1$. Could I obtain a continuous map $\gamma:S^n \rightarrow R$ from this?

Jean-Baptiste Vienney (Jan 26 2025 at 21:18):

Oh, maybe $\gamma(x_1,\dots,x_n)=i(x_1)$ for instance.

Jean-Baptiste Vienney (Jan 26 2025 at 21:20):

I can use the projections $p_i:S^n \rightarrow [0,1]$ for $1 \le i \le n$ to get continuous maps $\gamma_i=i \circ p_i:S^n \rightarrow [0,1]$.

Jean-Baptiste Vienney (Jan 26 2025 at 21:27):

I didn't even know precisely the definition of $\pi_n(R,0)$. Let's write it here.

Jean-Baptiste Vienney (Jan 26 2025 at 21:28):

First we have to choose a base point $a$ in $S^n$. Let's say $a=(1,0,\dots,0)$.

Jean-Baptiste Vienney (Jan 26 2025 at 21:33):

Now $\pi_n(R,0)$ is the set of homotopy classes of continuous maps $\sigma$ from $S^n$ to $R$ such that $\sigma(a)=0$.

Jean-Baptiste Vienney (Jan 26 2025 at 21:34):

Let $\sigma:S^n \rightarrow R$ be a continuous map such that $\sigma(a)=0$.

Jean-Baptiste Vienney (Jan 26 2025 at 21:35):

We want to build a continuous map $H:S^n \times [0,1] \rightarrow R$ such that $H(x_1,\dots,x_n,0)=\sigma(x_1,\dots,x_n)$ and $H(x_1,\dots,x_n,1)=0$ and $H(a,t)=0$.

Jean-Baptiste Vienney (Jan 26 2025 at 21:37):

Random try: maybe take $H(x_1,\dots,x_n,t)=i(t)\sigma(x_1,\dots,x_n)$.

Jean-Baptiste Vienney (Jan 26 2025 at 21:39):

Ok, it works.

Jean-Baptiste Vienney (Jan 26 2025 at 21:42):

Now I guess that all the $\pi_n(R,r)$ are isomorphic by translation.

Jean-Baptiste Vienney (Jan 26 2025 at 21:50):

Jean-Baptiste Vienney (Jan 26 2025 at 21:51):

By the way, something else:

• If $R$ is a topological ring then every path $\gamma$ from $0$ to $1$ in $R$ gives a continuous map $f$ from $\mathbb{R}$ to $R$ (I think).

Jean-Baptiste Vienney (Jan 26 2025 at 21:53):

Define $f(x)=\lfloor x \rfloor.1 + \gamma(x-\lfloor x \rfloor)$.

Jean-Baptiste Vienney (Jan 26 2025 at 21:54):

So that having a continuous path from $0$ to $1$ in $R$ would be equivalent to having a continuous map from $\mathbb{R}$ to $R$ which preserves $0$ and $1$.

Jean-Baptiste Vienney (Jan 26 2025 at 22:04):

It would be nice if a topological ring $R$ had a path from $0$ to $1$ iff it had a homomorphism of topological rings from $\mathbb{R}$ to its center... (i.e. if it was a topological $\mathbb{R}$-algebra)

Jean-Baptiste Vienney (Jan 26 2025 at 22:17):

Note also, that:

• if $R$ is a topological ring with $0 \neq 1$, if there is a path $\gamma$ from $0$ to $1$ in $R$ then $R$ is not totally disconnected since $\gamma([0,1])$ is a connected subspace of cardinal $>1$.

Jean-Baptiste Vienney (Jan 26 2025 at 23:29):

John Baez said:

And I almost feel like conjecturing that any topological ring is weakly homotopy equivalent to a discrete space! Certainly Tom showed any compact Hausdorff topological ring has this property - because they are totally disconnected. We have topological rings like $\mathbb{R}$ and $\mathbb{C}$ and $\mathbb{R}[x]$ and $\mathbb{R}[x,y]/\langle x^2 - 3y \rangle$ that are not totally disconnected, but they're contractible, so they're weakly homotopy equivalent to a point.

I think that the fundamental groupoid of the topological ring of continuous maps from $\mathbb{R}$ to $\mathbb{R}$ (say with the product topology) is not equivalent to a discrete category! (or is it… ??)

Kevin Carlson (Jan 26 2025 at 23:51):

Well, why do you think that?

Jean-Baptiste Vienney (Jan 26 2025 at 23:58):

Sorry, I’m wrong. The fundamental groupoid of every topological $\mathbb{R}$-algebra is equivalent to a discrete category.

Jean-Baptiste Vienney (Jan 27 2025 at 00:00):

Because if you have a loop $\gamma$ from $0$ to $0$ then you have the following homotopy from the constant loop equal to $0$ to $\gamma$: $H(s,t)=t\gamma(s)$.

Jean-Baptiste Vienney (Jan 27 2025 at 00:02):

So the fundamental group based at $0$ is trivial. By translation the fundamental group based at any point is trivial. It follows that the fundamental groupoid is thin.

Jean-Baptiste Vienney (Jan 27 2025 at 00:13):

To sum up we know that these three classes of topological rings have fundamental groupoid equivalent to a discrete category:

• the topological $\mathbb{R}$-algebras
• the topological fields
• the compact topological rings

Jean-Baptiste Vienney (Jan 27 2025 at 00:15):

Maybe we can generalize the first two cases to the case of a topological algebra over a topological field

Jean-Baptiste Vienney (Jan 27 2025 at 00:16):

I think this is not so much being a field which matters but having a field sitting inside the ring.

Jean-Baptiste Vienney (Jan 27 2025 at 00:26):

Brr sorry, the case of a topological algebra over a topological field follows from the case when you have a path from $0$ to $1$ exactly in the same way as for a topological field. (I think)

Jean-Baptiste Vienney (Jan 27 2025 at 00:31):

Now, I’m no longer really sure about topological algebras over a topological field.

Jean-Baptiste Vienney (Jan 27 2025 at 00:32):

If the field is not $\mathbb{R}$, I don’t know what to do.

John Baez (Jan 27 2025 at 05:26):

I think I see how to prove a dichotomy theorem: every topological field $F$ is either contractible or there is no (continuous) path from $x$ to $y$ when $x \ne y$.

Proof. If there's a path from $x$ to $y$ with $x \ne y$, then we can subtract $x$ from this path and get a path from $0$ to $y - x$. Then we can divide that path by $y - x$ and get a path $\gamma$ from $0$ to $1$.

Then, given a path $\gamma$ from $0$ to $1$, multiplication by $1 - \gamma(t)$ gives a homotopy from the identity map $\text{id}: F \to F$ to the map sending everything in $F$ to zero. Thus $R$ is contractible!

John Baez (Jan 27 2025 at 05:28):

Now I'll speed up a bit and leave out details. It follows from the dichotomy theorem that there are two cases: either $F$ is contractible or $F$ is weakly homotopy equivalent to the discrete space with one component for each point in $F$. Since a contractible space is homotopy equivalent to a discrete space, in either case $F$ is weakly homotopy equivalent to a discrete space!

John Baez (Jan 27 2025 at 05:30):

This means that topological fields are 'boring' from a homotopy theorist's point of view - but there are two different ways for them to be boring.

Todd Trimble (Feb 09 2025 at 16:23):

I had been away from Zulip for a while, and I've just skimmed over this thread.

Here I think is one way to produce topological rings with nontrivial fundamental groups. Please let me know if you see any problems. It proceeds through the following steps.

• The functor $B: \mathsf{Mon} \to \mathsf{Cat}$, sending a monoid $M$ to the obvious 1-object category $BM$, is product-preserving. Being product-preserving, it sends abelian group objects in $\mathsf{Mon}$ (which are the same as monoid objects in $(\mathsf{Ab}, \times)$, which are the same as abelian groups) to abelian group objects in $\mathsf{Cat}$. So for example, $A = B(\mathbb{Z}/2)$ is an abelian group object in $\mathsf{Cat}$.

• $\mathsf{Cat}$ is cartesian closed. Given an abelian group object $A$ in a cartesian closed category, its internal endo-hom object $[A, A]$ carries a ring object structure, with addition being defined "internally pointwise", and multiplication given by internal composition. (For $A = B(\mathbb{Z}/2)$, the underlying category of $[A, A]$ is a finite category; this will be needed later.)

• The nerve functor $N: \mathsf{Cat} \to \mathsf{Set}^{\Delta^{op}}$ and the geometric realization functor $R: \mathsf{Set}^{\Delta^{op}} \to \mathsf{Top}$ are both product-preserving. Here we need to be a little bit careful: for now, $\mathsf{Top}$ denotes a [[convenient category of topological spaces]], such as the category $\mathrm{CGWH}$ of compactly generated weakly Hausdorff spaces. Let's put a pin through that; I'll come back to this technical point in the next bullet point. In any case, the functor $RN: \mathsf{Cat} \to \mathsf{Top}$, being product-preserving, takes ring objects in $\mathsf{Cat}$ to ring objects in $\mathsf{Top}$, hence $\mathrm{CGWH}$-rings under one choice of notion of "convenient space".

• A $\mathrm{CGWH}$-ring might not be a topological ring since products $X \times_k Y$ in $\mathrm{CGWH}$ do not always match products $X \times Y$ in topological spaces. But under certain circumstances, they do match. For this, I'll point to Theorem A.6 in Hatcher's Algebraic Topology (p. 524 of https://pi.math.cornell.edu/~hatcher/AT/AT.pdf), where if $X$ and $Y$ are CW-complexes with countably many cells, then the canonical continuous map $X \times_k Y \to X \times Y$ is a homeomorphism. This applies in particular if $X = Y$ is the geometric realization of the nerve of a finite category, such as $[A, A]$ above.

• We may now deduce that $RN([A, A])$, with notation as above, is a topological ring. I claim this has nontrivial fundamental group (it's connected, by the way). This is because $A = [1, A]$ is a retract of $[A, A]$ (since $1$ is a retract of $A$). If $\pi_1(RN([A, A]))$ were trivial, then so would be its retract $\pi_1(RN A) = \pi_1(RNB(\mathbb{Z}/2))$. But $RNB(\mathbb{Z}/2)$ is a classifying space for the group $\mathbb{Z}/2$; its fundamental group is $\mathbb{Z}/2$, and we have reached a contradiction.

• Hence $RN([A, A])$ is a topological ring with nontrivial fundamental group.

Todd Trimble (Feb 09 2025 at 17:10):

Although now, I'm having some critical doubts. If there's a path $\alpha: I \to X$ in a topological ring, from the additive identity element $0 \in X$ to the multiplicative identity $1 \in X$, then you'd think there was a contracting homotopy $h: X \times I \to X$ defined by $h(x, t) = x\alpha(t)$.

I off-handedly said above that my construction is connected (but without giving the reason, which is that I thought that $[A, A]$ is connected as a category, hence its classifying space $RN[A, A]$ is also connected). This would imply that $RN[A, A]$ is path-connected (since CW-complexes are locally path-connected, so connected CW-complexes are path-connected), hence there should be a path from $0$ to $1$. So I seem to be pretty confused at the moment.

The reasoning in this comment seems a lot simpler than the reasoning in the previous comment, so I'm inclined to trust this comment more, but I don't see where the previous comment breaks down.

John Baez (Feb 09 2025 at 19:06):

Todd Trimble said:

Although now, I'm having some critical doubts. If there's a path $\alpha: I \to X$ in a topological ring, from the additive identity element $0 \in X$ to the multiplicative identity $1 \in X$, then you'd think there was a contracting homotopy $h: X \times I \to X$ defined by $h(x, t) = x\alpha(t)$.

That sounds right. I used a similar argument above to argue that in a topological field, either each point is its own path-connected component or the topological field is contractible. But without using division, you can use your argument to show that if there's a path from $0$ to $1$ in a topological ring, it must be contractible.

John Baez (Feb 09 2025 at 19:08):

I'm not seeing how this argument applies to your example, though!

Todd Trimble (Feb 09 2025 at 21:36):

One doofus error I made in my head was not using $[A, A]$ for the internal abelian group hom, but just the internal hom.

Forming the internal abelian group hom requires that the ambient cartesian closed category has equalizers. In other words, you'd want to internally form what in sets would be written as $\{f: A \to A|\; f(a + b) = f(a) + f(b)\}$, involving an equation. The left side of the equation $f(a + b)$ would correspond to a map

$[+, A]: [A, A] \to [A \times A, A]$

that internally sends $f: A \to A$ to the map $(a, b) \mapsto f(a + b)$. The right side of the equation $f(a) + f(b)$ corresponds to a composite

$[A, A] \overset{\mathrm{copy}}{\longrightarrow} [A \times A, A \times A] \overset{[1, +]}{\longrightarrow} [A \times A, A]$

where the first map internally sends $f: A \to A$ to the map $(a, b) \mapsto (f(a), f(b))$, and the second augments that map like so: $(a, b) \mapsto (f(a), f(b)) \mapsto f(a) + f(b)$. The equalizer of these two maps

$[A, A] \rightrightarrows [A \times A, A]$

is the internal object of abelian group homomorphisms $A \to A$. Let's call that equalizer $\mathsf{Ab}(A, A)$.

It's rather less clear from this point of view that $A$ is a retract of $\mathsf{Ab}(A, A)$, isn't it? In fact that looks downright dubious.

In which case, it's back to the old drawing board, to quote Marvin the Martian of Looney Tunes cartoon fame.