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For an ordinary functor $F\colon \mathcal{C} \to \mathcal{D}$ of categories, there is a construction $\operatorname{im} F$, the image of $F$, which is again a category, and $F$ factors through that image.

Is there anything vaguely like the image of a strong monoidal functor, which should be monoidal again?

What certainly doesn't work straightforwardly is just taking the image of the underlying functor and somehow putting a monoidal structure on it. Given $X, Y\colon \mathcal{C}$, how would you define $FX \otimes FY$? If you take the tensor product in $\mathcal{D}$, the result might not be in the image of $F$, but neither can you define it to be $F(X \otimes Y)$, since $F$ might no be injective on objects (so you can't find out a unique object to start with, and there is no canonical way to choose).

I'm utterly surprised I've never encountered such a construction. The only thing I've come across is for the case of fusion categories. There, the full subcategory spanned by summands of objects in the image of $F$ can be defined.

The only thing I can come up with for plain monoidal categories is the category spanned by the image of the underlying functor and all isomorphic objects. But this doesn't have good properties if I want to extend structures on $\mathcal{C}$ to $\mathcal{D}$. For example, if I have a braiding $c_{X,Y}\colon X \otimes Y \to Y \otimes X$ on $\mathcal{C}$, I can transport it onto objects of the form $FX$, but not onto all objects isomorphic to an $FX$, since I don't know which isomorphism to transport it along.

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  • $\begingroup$ Reading this and this has me convinced that the right way to think of the image is "having the objects of $\mathcal C$ and morphisms of $\mathcal D$. I think the following definition works: $\endgroup$ – Oscar Cunningham Sep 6 '16 at 19:51
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    $\begingroup$ The monoidal category $\mathrm{1im}\;F$ has the same objects as $\mathcal C$ and its morphisms $A\rightarrow B$ are $\mathcal D(FA,FB)$. Composition is as defined in $\mathcal D$. Define $\otimes$ as in $\mathcal C$ for objects. On morphisms, use the $\otimes$ in $\mathcal D$ to get something in $\mathcal D(FA\otimes FC,FB\otimes FD)$ and then use the coherence maps of $F$ to translate this into $\mathcal D(F(A\otimes C),F(B\otimes D))$. Then the canonical functors $\mathcal C\rightarrow \mathrm{1im}\;F\rightarrow\mathcal D$ are eso and full-and-faithfull respectively, which is what we want. $\endgroup$ – Oscar Cunningham Sep 6 '16 at 19:51
  • $\begingroup$ In higher category theory there are different notions of image living at different categorical levels (ncatlab.org/nlab/show/n-image). Functors between categories have 2 notions of image, neither of which is the one you've given (which I believe fails to be invariant under equivalences). Monoidal functors should have 3 because monoidal categories live in a 3-category. $\endgroup$ – Qiaochu Yuan Sep 6 '16 at 19:54
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    $\begingroup$ @QiaochuYuan Surely two of those three images must be the same. After all monoid homomorphisms don't have two notions of image (or do they?). $\endgroup$ – Oscar Cunningham Sep 6 '16 at 20:03
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    $\begingroup$ @Turion As Qiaochu says, there are definitely two notions of image of a functor $F:\mathcal C\rightarrow D$. The "$2$-image" is defined to be the category $\mathcal C$ with two morphisms quotiented together if they have the same image under $F$. The "$1$-image" is defined to be the full subcategory of $\mathcal D$ spanned by the objects of the form $FX$. I gave what I thought was the generalisation of the $1$-image (though Qiaochu disagrees). By "image" do you instead mean the $2$-image or something else? $\endgroup$ – Oscar Cunningham Sep 6 '16 at 20:34
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Let $F : \mathcal{C} \to \mathcal{D}$ be a strong monoidal functor. Since monoidal categories are non-strict structures in the sense that e.g. associativity of $\otimes$ only holds up to isomorphism, it makes sense to look at those objects of $\mathcal{D}$ which are isomorphic to objects of the form $F(X)$, where $X$ is an object of $\mathcal{C}$. These constitute a full subcategory of $\mathcal{D}$, which contains the unit object and is closed under tensor products, hence inherits a monoidal structure.

If (only) $\mathcal{C}$ has a braiding, there is no reason to expect that it induces a braiding on the image, since this image depends on $F$. For this, we should require that $\mathcal{D}$ is also braided and that $F$ is a braided monoidal functor. It follows that the image, as defined above, carries a braiding induced by $\mathcal{D}$, and that $F$ factors as a braided monoidal functor into it.

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