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No, this is not the case. Let the category C be vector spaces (say over the real numbers). Given any real number we get a natural transformation of the identity functor on C. On components for a giv …

I have not checked that your construction in invariant under (2-)equivalence of (coherent) 2-group. If it is not invariant, then your question is ill-posed. Assuming your question is well-posed and …

This is probably an easy question. Let C be a category with (finite) products. An internal hom in C category is an object uhom(X, Z) which represents the functor: Y |-----> hom(Y x X, Z) here "u …

This question has been bugging me since it was posted, in part because I keep thinking you should have a homotopy action and should take the homotopy quotient. But anyway.... whew! I have an example …

I will focus on the strictly injective case. Claim: The only strictly injective categories are the posets which are complete lattices. The strict injective property requires that you have the lifting …

This isn't really a fulll answer, but there is a obvious guess for (2). Consider those categories for which there exists a cardinal $\kappa$ and a set of $\kappa$-compact objects such that every objec …

Here is an example of how one might have stumbled upon the definition of a map of adjunctions. Suppose that you are working on a research project with a collaborator. Let's call her Jane for the sake …

[Note: This answer pertains to the orignal version of the question, and no longer applies given the subsequent edits]. To be a weak factorization system you need to factor morphisms as composites $ …

Let T be the algebraic theory of vector spaces (in the sense of Lawvere). Then a T-algebra in the category of small categories is a Baez-Crans 2-vector space. You seem to be asking about pseudo-T-alge …

I assume you meant "symmetric monoidal functors". Yes, this seems to hold. By your description you have constructed maps: $$ \eta: \mathcal{C} \cong \Omega B \mathcal{C}$$ $$ \varepsilon: B \Omega \ …

You can always do this. Take any $b$ and define $d = p_2 b$. Then $b' = (p_1, d)$ is equivalent to the original $b$. To see this note that $$(p_1, m) \circ b = (p_1b, m \circ (p_1 b, d)) \simeq id = …

For a category enriched in topological spaces, the usual classifying space can be made to take into account the topology on morphisms. More generally this works for categories internal to top and is d …

[Ignore this first part, I'm just leaving it for the context to the comments below.] It is hard for me to understand why you would want to enrich in symmetric monoidal categories, have an identity, a …

This is not true. It fails for essentially the same reason that the category of topological commutative groups fail to be an abelian category. For simplicity let's work over an algebraically closed …

A groupoid is a category in which all morphisms are invertible.(*) The groupoids form a very nice subclass of categories. The inclusion of the groupoids into the 2-category of small categories admits …