When does a functor F extend to a “localized” domain?

Suppose $\mathcal{C}, \mathcal{D}$ and $\mathcal{S}$ are symmetric monoidal categories. Let $F: \mathcal{C} \to \mathcal{D}$ be a functor and $S \subset \mathcal{C}$ such that every morphism in $S$ is an isomorphism. Then, it is possible to define $S^{-1} \mathcal{C}$. My question is what is required of $S$ in order for $F$ to extend to $S^{-1} \mathcal{C} \to \mathcal{D}$?

I am reading about this in Weibel's book on Algebraic K-theory, but cannot find anything about why $F$ can extend.

Intuitively, it would be a sort of generalization to how for $S$ a multiplicatively closed subset of a ring $R_1$, a map of rings $f: R_1 \to R_2$ extends to $f: S^{-1} R_1 \to R_2$ if and only if $f(S)$ consists of units in $R_2$.

• I don't understand your requirement that every morphism in $S$ is an isomorphism; in this case there's nothing to invert. The universal property of the localization is exactly the same as in ring theory: it's necessary and sufficient that $F$ send every morphism in $S$ to an isomorphism. – Qiaochu Yuan May 2 '17 at 17:13

As Qiaochu mentioned, it's exactly backwards to assume that $S$ consists of isomorphisms, although it's an inoffensive assumption to suppose that $S$ contains all the isomorphisms. I'd be surprised if Weibel really says nothing about when functors extend, because the definition of $S^{-1}\mathcal{C}$ is (or should be) that, for every $\mathcal{D}$, composing with the localization functor $\mathcal{C}\to S^{-1}\mathcal{C}$ gives an isomorphism between the category of functors $S^{-1}\mathcal{C}\to \mathcal{D}$ and the category of functors $\mathcal{C}\to\mathcal{D}$ such that $S$ lands in the isomorphisms of $\mathcal{D}$. Anything about strings of words built out of formal inverses of elements of $S$ is just by way of a construction, and a number of different constructions are important and useful in various contexts.
• The definition he gives is for a symmetric monoidal category $S$:"If S acts upon X, the category ⟨S,X⟩ has the same objects as X. A morphism from x to y in ⟨S, X⟩ is an equivalence class of pairs $(s,\phi s \Box x \to y)$, where s ∈ S and $\phi$ is a morphism in X. Two pairs $(s, \phi )$ and $(s′, \phi′)$ are equivalent in case there is an isomorphism $s\cong s'$ identifying $\phi$ with $s' \Box 􏰁x \cong s \Box 􏰁x \to y$ We shall write $S^{-1} X$ for ⟨S, S × X⟩, where S acts on both factors of S × X." Is this a particular case of the definition you gave? – math1234567 May 4 '17 at 10:16
• Oh, no, that's not the same notion I was thinking of. I don't know anything about that construction, which I've never seen used, but it might not be too hard to straighten out. I'm having trouble straighten out your description, since there are two distinct notions being called $S$, and it's unclear how the localization depends on which one. In your question, it seems to depend on a class of arrows in $X$, while in your comment, it seems to depend on the symmetrical monoidal category acting on $X$. – Kevin Carlson May 4 '17 at 15:48