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I feel that this is true but I'm unable to prove it formally: a monoid $M$ is $\omega$-presentable in the category $M$-$\mathbf{Set}$. This is the category of $(X,\rho)$ where $\rho:M\times X\to X$ satisfies $(a\cdot b)\cdot x=a\cdot (b\cdot x)$ and $1\cdot x=x$. The explicit definition of finite presentability is given in the snippet below:

Definition. An object $K$ of a category $\mathcal{K}$ is called finitely presentable provided that its hom-functor

$$\operatorname{hom}(K, {-}) : \mathcal{K} \to \mathbf{Set}$$

preserves directed colimits.

Explicitly: $K$ is finitely presentable iff for each directed diagram

$$D : (I, \le) \to \mathcal{K},$$

each colimit cocone $D_i \overset{c_i}{\rightarrow} C ~ (= \operatorname{colim} D), i \in I$, and each morphism $f : K \to C$ there exists $i$ such that

(1) $f$ factorizes through $c_i$, i.e., $f = c_i \cdot g ~ (i \in I)$ for some $g : K \to D_i$,

and

(2) the factorization is essentially unique in the sense that if $f = c_i \cdot g = c_i \cdot g'$, then $D(i \rightarrow j) \cdot g = D(i \rightarrow j) \cdot g'$ for some $j \ge i$.

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  • $\begingroup$ Hyphen, not minus sign. $\endgroup$ Apr 27, 2023 at 18:26
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    $\begingroup$ OK, first step would be to understand how the hom-functor $\operatorname{hom}_{M{-}\mathbf{Set}}(M, {-})$ acts. $\endgroup$ Apr 27, 2023 at 18:44
  • $\begingroup$ @DanielSchepler A far too complex thing for me :-( $\endgroup$
    – user122424
    Apr 28, 2023 at 16:12
  • $\begingroup$ In fact, $\operatorname{hom}_{M{-}\mathbf{Set}}(M, {-})$ is isomorphic to the underlying set functor $M{-}\mathbf{Set} \to \mathbf{Set}$. The basic idea is that an $M$-equivariant morphism from $M$ to an $M$-set $S$ is uniquely determined by where it sends the identity of $M$, and conversely, for any element $s$ of the underlying set of $S$, there is a morphism $M \to S$ defined by $m \mapsto m \cdot s$. (You can also think of this as saying that $M$ is "the free $M$-set on one generator".) $\endgroup$ Apr 28, 2023 at 16:28
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    $\begingroup$ And for your other question, the free $M$-set on 2 generators looks like $M \sqcup M$ (a disjoint union of two copies of $M$) where the action of $M$ is: $m \cdot left(m') = left(m m'), m \cdot right(m') = right(m m')$ (where $left$ and $right$ are the two maps $M \to M \sqcup M$). $\endgroup$ Apr 28, 2023 at 17:12

1 Answer 1

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Let me first review a couple basic properties of directed limits of $M$-sets:

Lemma 1: Suppose $x \in \varinjlim D(i)$. Then there exists some $i \in I$ and $x_i \in D(i)$ such that $x = c_i(x_i)$.

Proof: Consider $S := \bigcup_{i\in I} \operatorname{im}(c_i)$. It is straightforward to show that this is a sub-$M$-set of $\varinjlim D(i)$; and by construction, each $c_i$ factors (uniquely) through $S$. It follows that the factors of $c_i$ induce a morphism of $M$-sets $\varinjlim D(i) \to S$ such that composing with the inclusion map of $S$ gives the identity map on $\varinjlim D(i)$. This implies that $S$ is all of $\varinjlim D(i)$. $\square$

Lemma 2: Suppose $x_i \in D(i)$, $y_j \in D(j)$ satisfy $c_i(x_i) = c_j(y_j)$. Then there exists $k \ge i, j$ such that $D(i\to k)(x_i) = D(j\to k)(y_j)$ in $D(k)$.

Proof: This one is a bit trickier. One way to see this is to construct an $M$-set of so-called germs of functions. Here, an element is given by some $i \in I$ and a function $f \in \prod_{j\ge i} D(j)$ -- i.e. $f$ is a function with domain $\{ j \in I \mid j \ge i \}$ such that for each $j\ge i$, we have $f(j) \in D(j)$. However, we need to give a condition that two germs $f \in \prod_{j\ge i} D(j)$ and $f' \in \prod_{j\ge i'} D(j)$ are considered equal if there exists $i'' \ge i, i'$ such that $f(j) = f'(j)$ whenever $j \ge i''$. (The proof that this gives an equivalence relation on "pre-germs" should be straightforward.) We can define an action of $M$ on the set of pre-germs of functions, where $m \cdot f$ has the same domain as $f$ and $(m \cdot f)(j) := m \cdot (f(j))$. (Check that this respects the equivalence relation, and therefore induces an $M$-set structure on the set $G$ of germs of functions.)

Now for each $i$, we can define a morphism of $M$-sets $D(i) \to G$ where $x \in D(i)$ maps to the pre-germ with domain $\{ j \in I \mid j \ge i \}$ and function $j \mapsto D(i\to j)(x)$, then to the corresponding germ. It should be a straightforward exercise to show that this forms a compatible collection of morphisms $D(i) \to G$, and therefore induces a morphism $\varinjlim D(i) \to G$.

Now, suppose we have $x_i \in D(i), y_{i'} \in D(j)$ such that whenever $j \ge i, i'$, we have $D(i\to j)(x_i) \ne D(i'\to j)(y_{i'})$. This would imply that the germs $j \mapsto D(i\to j)(x_i)$ and $j \mapsto D(i'\to j)(y_{i'})$ generated by $x_i$ and $y_{i'}$ respectively are unequal in $G$. Therefore, we must have $c_i(x_i) \ne c_{i'}(y_{i'})$ in $\varinjlim D(i)$. $\square$


Note that an alternative proof for both lemmas would be to exploit the usual construction of a directed colimit of $M$-sets: start with $\bigsqcup_{i\in I} (\{ i \} \times D(i))$. Then quotient by the equivalence relation where $(i, x_i) \sim (j, y_j)$ if there is some $k \ge i, j$ with $D(i\to k)(x_i) = D(j\to k)(y_j)$ in $D(k)$. Show that this forms an $M$-set with action of $M$ induced by the action $m \cdot (i, x_i) := (i, m \cdot x_i)$. Show that the canonical morphisms $D(i)$ to the quotient form a compatible collection, and show that this system satisfies the required universal property of the directed colimit. Now, note that under this construction, both lemmas are pretty much immediate.


Now, finally, to show directly that the explicit conditions are satisfied:

(1) Suppose we have a morphism of $M$-sets $f : M \to \varinjlim D(i)$. Then by lemma 1, there exists some $i \in I$ and $x_i \in D(i)$ such that $f(e) = c_i(x_i)$. Now define $g : M \to D(i)$ by $g(m) := m \cdot x_i$. It is straightforward to check that $g$ is a morphism of $M$-sets. Also, for any $m \in M$, $$f(m) = f(m\cdot e) = m \cdot f(e) = m \cdot c_i(x_i) = c_i(m\cdot x_i) = c_i(g(m)),$$ so $f = c_i \circ g$ as required.

(2) Suppose we have two morphisms of $M$-sets $g, g' : M \to D(i)$ such that $c_i \circ g = c_i \circ g'$. Then by lemma 2, there exists some $j\ge i$ such that $D(i\to j)(g(e)) = D(i\to j)(g'(e))$. This implies that for any $m\in M$, $$D(i\to j)(g(m)) = D(i\to j)(g(m\cdot e)) = m \cdot D(i\to j)(g(e)) = m \cdot D(i\to j)(g'(e)) \\ = D(i\to j)(g'(m\cdot e)) = D(i\to j)(g'(m)),$$ so $D(i\to j) \circ g = D(i\to j) \circ g'$ as required.

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  • $\begingroup$ I'm now studying your pretty long answer. I may have more questions later, but for now: why each $c_i$ factors uniquely through $S$ ? $\endgroup$
    – user122424
    Apr 29, 2023 at 14:35
  • $\begingroup$ Because $\operatorname{im}(c_i) \subseteq S$. $\endgroup$ Apr 30, 2023 at 15:39
  • $\begingroup$ Right. And why the composition with inclusion gives the identity on $\varinjlim D(i)$ and how it follows that $S$ is all of $\varinjlim D(i)$ ? $\endgroup$
    – user122424
    Apr 30, 2023 at 16:49
  • $\begingroup$ Well, if $h$ is the induced morphism $\varinjlim D(i) \to S$, then by construction $h \circ c_i$ is the factor of $c_i$ through $S$, so $\operatorname{inc}_S \circ h \circ c_i = c_i = \operatorname{id} \circ c_i$. Therefore, by the uniqueness part of the universal property of $\varinjlim D(i)$, $\operatorname{inc}_S \circ h = \operatorname{id}$. $\endgroup$ May 1, 2023 at 16:44
  • $\begingroup$ And then, for every $x \in \varinjlim D(i)$, we have $x = \operatorname{inc}_S(h(x)) \in S$. $\endgroup$ May 1, 2023 at 16:45

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