4
$\begingroup$

The $n$Lab writes (prop. 2.2 in https://ncatlab.org/nlab/show/category+of+monoids) :

Let $C$ be a monoidal category with countable coproducts that are preserved by the tensor product. Then the forgetful functor $U_C$ has a left adjoint $F_C:C\rightarrow Mon(C)$.

Suppose now $C$ is a symmetric monoidal category. Under what conditions does the canonical forgetful functor $U_C:CMon(C)\rightarrow C$ from the category of commutative monoid objects in $C$ have a left adjoint? How is this left adjoint constructed?

$\endgroup$

1 Answer 1

6
$\begingroup$

The left adjoint is the functor Sym, defined as $Sym(X) = I \coprod X \coprod (X \otimes X)/\Sigma_2 \coprod (X\otimes X\otimes X)/\Sigma_3 \coprod \dots$, where $I$ is the unit. I only ever work in closed, cocomplete settings, where this always exists, but in general I think it should exist under the same conditions as the free monoid functor, as long as the colimits $X^{\otimes n}/\Sigma_n$ exist.

$\endgroup$
2
  • $\begingroup$ Of course we should also require that $\otimes$ preserves coproducts and quotients in both variables (otherwise, how to define the monoid structure on this coproduct of quotients?). Usually one requires that $\otimes$ preseves colimits in both variables (which is automatic in the closed setting). $\endgroup$
    – HeinrichD
    Sep 12, 2016 at 8:01
  • 1
    $\begingroup$ A reference is Proposition 1.3.1 in Florian Marty's thesis "Des Ouverts Zariski et des Morphismes Lisses en Geometrie Relative". $\endgroup$
    – HeinrichD
    Sep 12, 2016 at 8:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .