Under which 'minimal' conditions on a symmetric monoidal category does an abelization functor from its monoids to its commutative monoids exist?

More precisely: let $(\mathcal{C},\otimes,1)$ be a symmetric monoidal category, and let $U : \mathrm{CMon}(\mathcal{C}) \rightarrow \mathrm{Mon}(\mathcal{C})$ be the forgetful functor from its category of commutative monoids to its category of monoids. Define the abelianization functor $(-)_{\mathrm{ab}}$, if it exists, as the left adjoint of the forgetful functor ($(-)_{\mathrm{ab}} \dashv U$).

It seems to me that this functor does not necessarily exist in general, and one has to assume some extra conditions on the symmetric monoidal category $\mathcal{C}$. Which 'minimal' such extra conditions do you know?

(Of course this 'minimal' is not well defined, as some conditions might not be weaker than others, just different, that's why I put it in quotes and by that I mean a general idea of 'as weak conditions as possible').

The page abelianization in nLab, under Section 3, mentions "we can do abelianization of monoid objects in many monoidal categories", the word many (and not all) suggesting that it is indeed not possible to do it in general, but it does not give any further hints on what kind of conditions one needs.


  • $\begingroup$ In full generality, the existence of left adjoints is linked to the existence of colimits; all in all, abelianisation will exist in every category of internal monoids in $C$ such that $U : Ab(C) \to Mon(C)$ satisfies the assumptions of the adjoint functor theorem. $\endgroup$
    – fosco
    Mar 21, 2021 at 12:02
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    $\begingroup$ An example where the abelianization functor does not exist: take $\mathcal{C}$ to be the category of sets whose cardinality is either 1 or infinite with the product monoidal structure. Infinite groups are examples of monoids in this category. There are infinite groups whose (standard) abelianization is a non-trivial finite group, and these won't exist as commutative monoid objects in $\mathcal{C}$. $\endgroup$ Mar 21, 2021 at 14:43

1 Answer 1


Here is an approach that has the advantage of avoiding to go in the details of a concrete construction. More explicit constructions of the abelianization might give even more general results, in terms of corollary 2 below they would corresponds to more explicit description of the object $X_{ab}$, but I think corollary 2 below already cover most of the applications.

Theorem: Let $C$ be a symmetric monoidal locally presentable category in which the tensor product is accessible. Then the categories of (commutative) monoids in $C$ is locally presentable.

Proof: The categories of monoids or commutative monoids can be expressed as Cat-enriched limits of powers of $C$ and map induced by the tensor product between them. So as long as the tensor product is accessible, these are accessible categories because Cat-enriched limits of accessible categories and accessible functors between are accessible. As these category have all limits (they are created by the forgetful functor to $C$) they are locally presentable categories.

Corollary 1: Let $C$ be a locally presentable symmetric monoidal category in which the tensor product is accessible. Then the abelianization functor exists.

Proof: This follows from the special adjoint functor theorem applied to the forget full functor from commutative monoids to moinods. Fix $\kappa$ such that $\otimes$ is $\kappa$-accessible. The forgetfull functor commutes to all limits and $\kappa$-directed colimits, i.e. it is an accessible right adjoint , so it has a left adjoint by the special adjoint functor theorem.

Exploiting this, one can do much better and drop the local presentability assumption by observing that "the universal example" is itself locally presentable:

Corollary 2: Let $C$ be a symmetric monoidal category with all colimits in which the tensor product preserves $\kappa$-directed colimits for some regular cardinal $\kappa$. Then the abelianization functor for $C$ exists.

Proof: Let $M_\kappa$ be the "free symetric monoidal category with all $\kappa$-small colimits generated by a monoid object $X \in M_\kappa$".

It exists, because all the structure I've put on $M_\kappa$ can be encoded as a $\kappa^+$ essentially algebraic theory (let says in a bicategorical sense for simplicity, though it would also work in a $1$-categorical sense).

Let $V_\kappa$ be the $\kappa$-ind completion of $M_\kappa$. Because Ind completion commutes to products, the tensor product on $M_\kappa$ induces a $\kappa$-accessible tensor product on $V_\kappa$, making $V_\kappa$ a symmetric monoidal category (if you prefer, ind completion being monoidal for the product, it sends monoids object to monoids object).

It follows that $V_\kappa$ is a locally presentable category with a $\kappa$-accessible tensor product, so one can apply the previous result and the universal monoid object $X \in M_\kappa$ admit a symmetrisation $X_{ab}$ in $V_\kappa$.

If now $C$ is a general symmetric monoidal category with all colimits and such that $\otimes$ preserves $\kappa$-directed colimits in each variables. Then given a monoid object $T \in C$ you have an essentially unique (strong) symmetric monoidal functor $F: M_\kappa \to C$ that preserves $\kappa$-small colimits and sends $X$ (as a monoid) to $T$.

It extends to a monoidal functor preserving all colimits $V_\kappa \to C$, in fact (as $V_\kappa$ is locally presentable), it is a left adjoint functor. It then follows from a general argument that the image of $X_{ab}$ by this functor is an abelianisation of $X$ (one just check the universal property in $C$ using the universal property of $X_{ab}$ in $V_\kappa$ and the adjunction).

  • $\begingroup$ Why are limits of monoids and commutative monoids created by the forgetful functor to C? (Assuming the stated generality of your answer.) $\endgroup$ Mar 21, 2021 at 16:09
  • $\begingroup$ Proposition in Lurie's Higher Algebra requires the tensor product to preserve countable coproducts in each variable. How do you show that the forgetful functor creates limits without assuming such a condition? $\endgroup$ Mar 21, 2021 at 16:22
  • $\begingroup$ @DmitriPavlov : this is a formal statement. Given a diagram of monoids having a limit in C, you get a monoid structure on the limit directly from the universal property and then you just check everything. If you want a formal argument there is a proof in Lurie Higher algebra in the context of $\infty$-operads (Corollary and I'm sure there are more classical reference in ordinary category theory though... $\endgroup$ Mar 21, 2021 at 16:23
  • $\begingroup$ I see, it appears that the condition of preservation of coproducts is simply not necessary for the first part of Proposition in Higher Algebra (existence of a left adjoint). $\endgroup$ Mar 21, 2021 at 16:30
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    $\begingroup$ @DmitriPavlov: Yes (as Simon said), the statement that limits of monoids and commutative monoids are created by the forgetful functor can be proved formally without any assumption on C. A more classical reference than Lurie's Higher algebra in ordinary category theory is - for example - the PhD thesis of Florian Marty (in the proof of Proposition 1.2.14) thesesups.ups-tlse.fr/540 $\endgroup$ May 31, 2021 at 11:03

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