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Given a functor $F : \mathcal C \to \mathcal C$ with initial algebra $\alpha : FA \to A$, and another algebra $\xi : FX \to X$, we obtain a unique morphism $\mathsf{fold}~\xi : A \to X$ such that $\mathsf{fold}~\xi \circ \alpha = \xi \circ F(\mathsf{fold}~\xi)$.

I am looking for a sufficient condition - ideally phrasable as $P(F) \wedge Q(\xi)$ rather than $R(F, \xi)$ - which guarantees that:

  1. $\mathsf{fold}~\xi$ is monomorphic,
  2. $\mathsf{fold}~\xi$ is epimorphic.

(I'm looking for separate conditions for 1 and for 2.)

I would expect:

  1. $F$ preserves monomorphism and $\xi$ is monomorphic
  2. $F$ preserves epimorphism and $\xi$ is epimorphic

but I can't seem to prove this.

EDIT: I'm also interested in the dual question (which, mathematically, of course just has dual answers). I mention this anyway, because Valery Isaev's answer uses local finite presentability, which is something that might not have been answered had I asked the dual question. So feel free to use concepts that are only dually familiar. In particular, I'm willing to assume that $F^{op}$ is polynomial (or similar in a less powerful category, e.g. $FX$ is a coproduct of powers of $X$ in $\mathcal C^{op}$).

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  • $\begingroup$ By "$Q(\xi)$" do you mean a property of the morphism $\xi$ purely as a morphism in $C$ without any reference to the fact that its domain happens to be of the form $FX$? Or would you include properties of $\xi$ as an $F$-algebra? $\endgroup$ – Mike Shulman Mar 10 at 14:13
  • $\begingroup$ Yes, I've realized the separation of concerns is a bit ill-typed. I'm willing to consider $\xi$ as an $F$-algebra. Anything is better than requiring that $0 \to X$ be epimorphic. $\endgroup$ – dremodaris Mar 11 at 11:43
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Suppose $({\cal E,M})$ is a factorization system on $\cal C$ such that $\cal M$ consists of monomorphisms and $F$ preserves $\cal E$-morphisms. If $X$ is an $F$-algebra, define an $\cal M$-subalgebra of $X$ to be a morphism $Y\to X$ of $F$-algebras that is an $\cal M$-morphism. Then if $A$ is the initial $F$-algebra, the map $A\to X$ lies in $\cal E$ if and only if $X$ has no proper $\cal M$-subalgebras.

In the "if" direction, note that the assumption that $F$ preserves $\cal E$-morphisms implies that $({\cal E,M})$-factorizations lift to $F$-algebras. In particular, if we factor the unique $F$-algebra morphism $A\to X$ as $A\to Z \to X$, then the unique lifting property of $F A \to F Z$ (which is in $\cal E$) against $Z\to X$ (which is in $\cal M$) makes $Z$ an $F$-algebra and the two maps $F$-algebra maps. In particular, $Z$ is an $\cal M$-subalgebra of $X$. Thus, if $X$ has no proper $\cal M$-subalgebras, then $Z\cong X$ and hence $A\to X$ is in $\cal E$.

Conversely, if $A\to X$ is in $\cal E$, suppose $Z\to X$ is an $\cal M$-subalgebra. Then $A\to X$ factors through $Z$ (since $A$ is the initial $F$-algebra), but then $Z\to X$ must be an isomorphism since it is an $\cal M$-morphism that is factored through by some $\cal E$-morphism.

In particular, if $\cal C$ admits (epi, strong mono) factorizations, then $A\to X$ is epi if and only if $X$ has no proper strong-subalgebras.

EDIT: Note that if $X$ has no proper $\cal M$-subalgebras, then $\xi : F X \to X$ is in $\cal E$. For we can $({\cal E,M})$-factor it as $F X \to Z\to X$ and show that $Z$ becomes an $\cal M$-subalgebra. The assumption on $X$ then implies $Z\cong X$, hence $\xi\in\cal E$.

On the other hand, as Valery pointed out, taking $F=\rm Id$ shows that the converse can't hold. Any endo-epimorphism $X\to X$ (e.g. $(+1) : \mathbb{Z}\to\mathbb{Z}$) gives an $F$-algebra $\xi$ that is an epimorphism, but $0 = A \to X$ will rarely be epi.

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If $\mathcal{C}$ has an initial object and the unique map $0 \to X$ is an epimorphism, then $\mathrm{fold}\ \xi : A \to X$ is an epimorphism. This follows from the fact that the composite $0 \to A \to X$ is an epimorphism. This condition seems too restrictive, but it is "almost necessary". Indeed, if $F(0) \simeq 0$ (for example, if $F$ is the identity functor), then $A \simeq 0$, so $\mathrm{fold}\ \xi : A \to X$ is an epimorphism if and only if $0 \to X$ is. It is hard to imagine some nice sufficient properties of $F$ which are not true for the identity functor.

Now, let's discuss when $\mathrm{fold}\ \xi$ is a monomorphism. Suppose that the following conditions hold:

  • $\mathcal{C}$ is locally finitely presentable.
  • $F$ preserves sufficiently long sequential colimits.
  • The unique map $0 \to X$ is a monomorphism.
  • The map $\xi : F(X) \to X$ is a monomorphism.
  • $F$ preserves monomorphisms.

Then $\mathrm{fold}\ \xi : A \to X$ is a monomorphism. Indeed, if the second condition holds, then $A$ can be constructed as a colimit of a transfinite sequence $$ 0 \to F(0) \to F^2(0) \to F^3(0) \to \ldots$$ This colimit exists since $\mathcal{C}$ is locally presentable. The map $f^n : F^n(0) \to X$ is constructed as the composite $F^n(0) \xrightarrow{F(f^{n-1})} F(X) \xrightarrow{\xi} X$. Since $0 \to X$ and $\xi$ are monomorphisms and $F$ preserves monomorphisms, we can prove that $f^n$ is a monomorphism by induction. At a limit stage, we need to use the fact that $\mathcal{C}$ is locally finitely presentable. Such a category has a generator consisting of finitely presentable objects. Thus, to prove that a map $f^\alpha : F^\alpha(0) \to X$ is a monomorphism, we just need to show that for every pair of maps $g,h : S \to F^\alpha(0)$, where $S$ is finitely presentable, if $f^\alpha \circ g = f^\alpha \circ h$, then $g = h$. Since $S$ is finitely presentable, maps $g$ and $h$ factor through $F^\beta(0)$ for some $\beta < \alpha$. Now, it follows that $g = h$ by induction hypothesis. This shows that $\mathrm{fold}\ \xi$ is a monomorphism since it equals $f^\lambda : F^\lambda(0) \to X$ for some $\lambda$.

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    $\begingroup$ I think in place of local finite presentability, it also suffices to assume $C$ is a Grothendieck topos, or more generally an exhaustive category (ncatlab.org/nlab/show/exhaustive+category), since then $F^\alpha(0)\to X$ is a union of a chain of subobjects of $X$ and hence also a subobject. $\endgroup$ – Mike Shulman Mar 10 at 4:33
  • $\begingroup$ However I'm not sure what you mean by "It is hard to imagine some nice sufficient properties of $F$ which are not true for the identity functor." There are plenty of endofunctors for which we want to consider initial algebras that don't preserve the initial object; indeed as you note, if $F0=0$ then $A=0$, so any functor with an interesting initial algebra must not preserve the initial object. For instance, $F(X) = X+1$, whose initial algebra is the natural numbers. $\endgroup$ – Mike Shulman Mar 10 at 4:35
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    $\begingroup$ @MikeShulman I mean a sufficient condition on $F$ cannot be of the form "$F$ preserves something", or "$F$ is a polynomial functor", or anything like that since the identity functor satisfies these conditions. Of course, we can consider conditions such as "$F$ does not preserve the initial object", but it seems less natural to me. $\endgroup$ – Valery Isaev Mar 10 at 11:45
  • $\begingroup$ Well of course you would have to put conditions on $X$ too. $\endgroup$ – Mike Shulman Mar 10 at 14:11

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