Let $T$ be a first order theory, $M$ its elementary class and consider its forgetfull functor $$M \stackrel{U}{\to} \text{Set}. $$

I would like to characterize those theories such that $U$ has a left adjoint. In fact this can be done.

If $M$ is cocomplete then $U$ has a left adjoint iff $T$ is equivalent to a limit theory in $L_{\omega}$.

Proof: This is quite an exercise. I will prove just the necessity. If $F$ exists then $F(\{\cdot\})$ is finitely presentable and also a strong generator of $M$ (here I am using that $U$ always commutes with directed colimits when T is first order), thus $M$ is finitely locally presentable. By the characterization theorem of finitely locally presentable categories, $T$ is equivalent to a limit theory.

(Q1): What if $M$ is not cocomplete?

In fact if the adjoint exist, $M$ must be finitely accessible but I do not know any characterization theorem for finitely accessible categories. This is anyway a strong request on the elementary class.

If U has a left adjoint then $M$ is finitely accessible.

Now there are two natural questions.

(Q2): Is the other implication true?

(Q3): What happens if one replaces $M$ with an abstract elementary class?

  • 2
    $\begingroup$ The morphisms of $M$ are elementary embeddings? Or do you allow all $\mathcal{L}$-homomorphisms? (in a relational language, do you intend weak homomorphisms or strong?) $\endgroup$ Jun 15, 2017 at 14:42
  • $\begingroup$ Thanks for the comment. I was thinking to $\mathcal{L}$-homomorphisms. But we can choose the environment that best settle the question. $\endgroup$ Jun 15, 2017 at 14:43
  • $\begingroup$ What is a limit theory? $\endgroup$ Jun 15, 2017 at 15:56
  • 1
    $\begingroup$ A limit sentence is a sentence of the following type: $$ (\forall x_1, \dots, x_n)(\phi( x_1, \dots, x_n) \Rightarrow (\exists ! y_1, \dots, y_m)\psi( x_1, \dots, x_n,y_1, \dots, y_m)).$$ Where $\phi$ and $\psi$ are conjunctions of atomic formulas. A set of limit sequences is a limit theory. Example: finitary essentially algebraic categories can be axiomatized by limit theories. You can find this definition in Locally Accessible and Presentable Categories, Adamek-Rosicky. $\endgroup$ Jun 15, 2017 at 16:01

2 Answers 2


Here's a second attempt at answering the "homomorphism" version of Q3.

Proposition: Suppose that $U: \mathrm{Hom}(T) \to \mathsf{Set}$ has a left adjoint $F$. Expand $T$ by those definable functions in $T$ which are preserved by all homomorphisms between models of $T$ to obtain a theory $T'$. Let $V(T')$ be the varietal hull of $T$, i.e. it has all the function symbols of $T'$ but not the relation symbols, and is axiomatized by the universally quantified equations that hold in $T'$. Then

  1. The reduct of $F$ to just the $T'$-function symbols is the usual free functor $\mathsf{Set} \to \mathrm{Hom}(V(T'))$ (this specifies the interpretation of the $T'$-function symbols in $F(n)$).
  2. If $\phi$ is an atomic formula in $T'$, then $\forall \bar x. \phi(\bar x)$ holds if and only if $\phi(\bar e)$ holds in $F(n)$, where $\bar e$ are the generators of $F(n)$ (this specifies the interpretation of the $T'$-relation symbols in $F(n)$).

Conversely, if $V$ is a variety, $T'$ is an expansion of $V$ with the same function symbols, and $T$ is a reduct of $T$ with the same relation symbols and $\mathrm{Hom}(T) = \mathrm{Hom}(T')$, then the free $V$-algebra functor lifts to a left adjoint to $U: \mathrm{Hom}(T') \to \mathsf{Set}$ if and only if the $\mathrm{Lang}(T')$-structure on $F(n)$ defined by (2) models $T'$.

Proof: Let us prove the first part. For $\bar a \in M \in \mathrm{Hom}(T)$, let $\langle \bar a \rangle: F(n) \to M$ denote the unique homomorphism such that $ \langle \bar a \rangle(\bar e) = \bar a$. For $t \in F(n)$, let $[t]_M (\bar a) = \langle \bar a \rangle(t)$. Let $\mathbf{R}(\bar x, \bar{\mathbf{y}}) = \wedge\wedge_i R_i(\bar x,\bar{\mathbf{y}})$ be the infinite conjunction of all positive quantifier-free formulas $R_i(\bar x, \bar{\mathbf{y}})$ such that $F(n) \models R_i(\bar e,\bar{\mathbf{t}})$, where $\bar{\mathbf{t}}$ is a fixed enumeration of the elements of $F(n)$. The uniqueness of the homomorphism $\langle \bar a \rangle$ for each $\bar a$ tells us that $T \vdash \forall \bar x,\bar{\mathbf{y}},\bar{\mathbf{y}}'. \mathbf{R}(\bar x,\bar{\mathbf{y}}) \wedge \mathbf{R}(\bar x, \bar{\mathbf{y}}') \to \wedge \wedge_i \mathbf{y}_i = \mathbf{y}'_i$. For any $t \in F(n)$, it follows by compactness that we have $T \vdash \forall \bar x, \bar y, \bar y'. R_t(\bar x,y,\bar y) \wedge R_t(\bar x,y',\bar y) \to y = y'$ for some (positive, quantifier-free) $R_t(\bar x, y, \bar y)$, and that in fact $\exists \bar y. R_t(\bar x, y,\bar y)$ is a definition of $y=[t](\bar x)$ valid in all models of $T$. So there is a function symbol $f(\bar x)$ in $T'$ such that $f(\bar a) = [t](\bar a)$ for all $\bar a$ and in particular, $f(\bar e) = [t](\bar e) = t$.

This tells us that every $t \in F(n)$ is representable as $t = f(\bar e)$ for some function $f(\bar x)$ in $T'$. Now if $f(\bar e) = g(\bar e)$ holds for two functions $f(\bar x)$ and $g(\bar x)$ in $T'$, then for any $\bar a \in M^n$ we have $f(\bar a) = f(\langle \bar a \rangle(\bar e)) = \langle \bar a \rangle(f(\bar e)) = \langle \bar a \rangle (g(\bar e)) = g(\langle \bar a \rangle(\bar e)) = g(\bar a)$, so the universal equation $\forall \bar x . f(\bar x) = g(\bar x)$ holds in $V(T')$. Conversely, of course, if $\forall \bar x. f(\bar x) = g(\bar x)$ holds, then in particular $f(\bar e) = g(\bar e)$ holds. This proves (1). Then (2) follows by similar reasoning.

(As a side note, given that $F$ exists, the functions of $T'$ are precisely those definable in $T$ by formulas of the form $f(\bar x) = y \leftrightarrow \exists \bar y . R(\bar x, y, \bar y)$ with $R$ a positive quantifier-free formula; we have shown one direction, while the other holds for any theory $T$.)

The converse is clear.

  • $\begingroup$ I accepted but I have some questions. I will use notation of "Locally presentable and accessible categories". If I understood correctly you say that for a given theory $T$ in a language $\Sigma$ you can find an other language $\Sigma'$ and some equational presentation $E$ such that Hom($T$) $\subset$ Struct($\Sigma'$, $E$). Is this correct? Do you have a reference for this? Then you say that if this free functor exist, it is in fact the functor you get from free algebras. The you say that it exists when contidion $2$ holds. $\endgroup$ Jul 10, 2017 at 15:47
  • $\begingroup$ Your $Struct(\Sigma',E)$ sounds like my $Hom(V(T'))$, the "varietal hull" of $T'$. To be precise, there is a forgetful functor $Hom(T) \to Hom(V(T'))$ which is concrete -- it preserves underlying sets and the interpretations of all function symbols. This functor forgets the data of relation symbols from $Hom(T)$, so I wouldn't say that $Hom(T) \subset Hom(V(T'))$. If you're asking for a reference on the varietal hull of a first-order theory, I don't know any. I'm not sure there's really much to say about it. I can flesh out those arguments a bit more. $\endgroup$
    – Tim Campion
    Jul 10, 2017 at 21:48
  • $\begingroup$ But the functor is also full. Isn't it? So, somehow, it's a subcategory. $\endgroup$ Jul 11, 2017 at 14:43
  • $\begingroup$ No problem! The functor $Hom(T') \to Hom(V(T'))$ is not necessarily full. Because the structure of a $V(T')$- model doesn't say anything about how to interpret the relation symbols of $T'$, it's possible that we have a $V(T')$-model $M$ which extends to a $T'$-model in two different ways -- call them $M_1,M_2$. Because they are different, there is $m \in M$ and a primitive $T'$-relation $R$ such that $R(m)$ holds in $M_1$ but not $M_2$ (or vice versa; if so, swap $M_1$ with $M_2$). Then the identity is a $V(T')$-homomorphism $M \to M$ which does not lift to a $T'$-homomorphism $M_1 \to M_2$. $\endgroup$
    – Tim Campion
    Jul 11, 2017 at 14:58
  • $\begingroup$ Do we have a natural example of this case, i.e. a first order theory such that Hom($T$) is not locally presentable but there are free objects? $\endgroup$ Jul 13, 2017 at 11:01

Let me explain why the question, raised in the comments, of what the morphisms of $M$ are, sheds a lot of light on this question. If $T$ is a first-order theory I will use $\mathrm{Mod}(T)$ to denote its category of models with elementary embeddings as morphisms, and $\mathrm{Hom}(T)$ to denote its category of models with homomorphisms as morphisms.

Observation: Let $M$ be a category and $U: M \to \mathsf{Set}$ a functor, and suppose that $U$ factors through the non-full subcategory $\mathsf{Set}_\mathrm{inj}$ of sets and injective maps. Suppose also that $U$ has a left adjoint. Then:

  • $U$ must actually factor through the subcategory $\{\emptyset,1\}$.
  • If moreover $U$ is faithful, then $M$ is a poset.

A proof will be given at the end. In particular, the hypotheses of both bullet points are satisfied in the following cases:

  1. $M= \mathrm{Mod}(T)$ and $U$ the usual forgetful functor.

  2. $M$ is $\mathsf{Set}_\mathrm{inj}$.

  3. $(M,U)$ is any AEC, as in (Q3).

This leads to the following conclusions:

  • By (1) if in Q1 we take $M$ to mean $\mathrm{Mod}(T)$, then the answer is "$F$ exists only in degenerate cases".

  • By (2), the answer to Q2 is: "No".

  • By (3), the answer to Q3 is: "If $(M,U)$ is an AEC, then $F$ exists only in degenerate cases."

Let us also observe that

  1. For any first-order theory $T$, there is a first-order theory $T'$ (the Morleyization of $T$) such that $\mathrm{Hom}(T') = \mathrm{Mod}(T)$, with the same forgetful functor to $\mathsf{Set}$.

  2. $\mathrm{Hom}(T)$ need not be accessible. For example, let $T$ be the theory $\exists x \exists y x \neq y$. Then $M$ is the category $\mathsf{Set}_{\geq 2}$ of sets of cardinality at least 2, which doesn't even have split idempotents.

  3. If $T$ is a complete theory with infinite models, then $\mathrm{Mod}(T)$ is never finitely accessible.

Now, if $M$ means $\mathrm{Hom}(T)$, we can conclude that

  • On account of (1) the answer to Q1 is "There are lots cases where $F$ does not exist".
  • On account of (2) and OP's observation that if $F$ exists then $M$ is finitely accessible, the answer to Q1 is "There is at least one case, and likely lots of cases, where $F$ does not exist".
  • (1),and (3), together again with OP's observation about finite accessibility, give yet another way to see the answer to Q1 is "there are lots of cases for which $F$ does not exist.

But Q1 is pretty vague. It's possible that what the OP really means is "Even if it's rare for $F$ to exist, can we identify any conditions more general than cocompleteness under which $F$ exists?". In that case, the answer seems to be: "In light of the above observations, and especially (2), probably one can say no more about this question for $M = \mathrm{Hom}(T)$ than one can say about the case where $M$ is an arbitrary concrete category (which is to say, not much)".

Proof of Observation:

In $\mathsf{Set}$, every morphism except for the morphisms $\emptyset \to S$ where $S \neq \emptyset$ factors as a split mono followed by a split epi. But in $M$, every split mono and every split epi is an isomorphism. So the restriction $F|_{\mathsf{Set}_{\neq \emptyset}}: \mathsf{Set}_{\neq \emptyset} \to M$ of $F$ to the category of nonempty sets factors through the localization of $\mathsf{Set}_{\neq \emptyset}$ at all morphisms, which is the terminal category.

Now the coproduct $2 = 1 \amalg 1$ is preserved by $F$. The two coproduct injections $1^\to_\to 2$ are exchanged by an automorphism of 2 which is sent to the identity by $F$, so the two coproduct injections are the same map $F(1) \to F(2)$. This implies that for any object $X \in M$, there is at most one map $F(1) \to X$. But by adjointness, maps $F(1) \to X$ correspond to elements of $X$. So every object of $M$ has at most one element.

Moreover, $F(\emptyset)$ must be an initial object. This, along with the universal property of $F(1)$, implies the first bullet point. If $U$ is faithful, this means that $M$ must be a poset, with $F(\emptyset)$ a bottom element $\bot$. By its universal property, $F(1) = \min(U^{-1}(1))$. Moreover no element of $U^{-1}(1)$ is below an element of $U^{-1}(0)$. Conversely, any poset and functor $U$ meeting this description have a left adjoint $F$ meeting this description.

  • $\begingroup$ This comment proves that one must look at morphisms and not just embeddings. $\endgroup$ Jul 2, 2017 at 14:20
  • $\begingroup$ @Ivan You seem unsatisfied, so I've rewritten my answer to clarify that it answers both of your less ambiguous questions Q2 and Q3, and I've also attempted to clarify in what sense it answers your ambiguous Q1. $\endgroup$
    – Tim Campion
    Jul 2, 2017 at 18:42
  • $\begingroup$ This is a beautiful edit to the comment, sure I will think about it and comment back. Thanks. $\endgroup$ Jul 3, 2017 at 14:03
  • $\begingroup$ I've changed my mind. The homomorphism version of the problem is not hard to answer; I've edited to answer it, unless I've made mistakes. But this class of theories doesn't look like a particularly promising one to study. $\endgroup$
    – Tim Campion
    Jul 4, 2017 at 2:09
  • $\begingroup$ Sorry, I think I made a mistake in that classification, so I've removed that edit. If I figure anything more out, I'll post it as a second answer $\endgroup$
    – Tim Campion
    Jul 6, 2017 at 20:23

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