Tag Info

New answers tagged

5

I hope what follows will clear up your confusion. A lax monoidal functor consists of the following data (satisfying some axioms that I won't spell out): One functor $\color{red}{F : \mathsf{C} \to \mathsf{D}}$. This means that for all objects $A \in \mathsf{C}$ you have an object $F(A) \in \mathsf{D}$, and for all morphisms $f : A \to B$ in $\mathsf{C}$ ...


1

The category of Banach spaces is locally $\aleph_1$-presentable.


2

To summarize the definition: an object of the "universal cover" $\hat{\mathcal{C}}$ of a category $\mathcal C$ is an indiscrete subcategory of $\mathcal{C}$, and a morphism $(A,(\phi_{a,a'})_{a,a' \in A}) \to (B,(\psi_{b,b'})_{b,b' \in B})$ is a collection $(f_{ab})_{a \in A,b \in B}$ of arrows which are natural in the obvious way. Composition is actually ...


2

The internal hom is defined via maps into it. You know nothing about the classification of maps into a coproduct, but you know how to classify the maps into a product, by the very definition of a product. You can also see how the product comes up in the following calculation: Let $A,B,C$ be graded $R$-modules (the case of chain complexes is similar, but ...


0

Yes. Have a look at: Lewis, G. (1972) Coherence for a closed functor. In: Mac Lane, S. (ed.) Coherence in Categories. Springer-Verlag Lecture Notes in Computer Science 281, 148–195 Epstein, D. B. A. (1966) Functors between tensored categories. Invent. Math. 1, 221–228


2

$\def\E{\mathbf E}\def\D{\mathbf D}$It's not counterintuitive at all when you understand how coends work! It takes a while but once you understand it, you understand it forever. According to the definition of coend as a universal cowedge, writing $\int^{e, e'}F(e,e')$ means that you are considering a functor $F\colon \E\times \E^{op}\times \E\times \E^{op} ...



Top 50 recent answers are included