# Tag Info

1

The standard simplicial enrichment on $Simp(C)$ is defined as follows. First, note that since $C$ is cocomplete, it can be considered to be tensored over $Set$: If $X$ is an object of $C$ and $S$ is a set, then $X\otimes S$ is a coproduct of copies of $X$ indexed by elements of the set $S$. If $K$ is a simplicial set and $A$ is an object of $Simp(C)$, we ...

3

For an exact functor $F$ between abelian categories, being faithful is equivalent to the condition $$FX=0\Rightarrow X=0$$ on objects, since a morphism is zero if and only if its image is zero, and exact functors preserve images. So, for your second question, the "criterion" proves this equivalent formulation of faithfulness.

0

I think this answers the first part of my question (before the edit): Suppose that the map $ev:X\otimes X^\vee \to \mathbf{1}$ is not an epimorphism. Then there exists an object $Z$ and distinct morphisms $\alpha, \beta: \mathbf{1}\to Z$ such that $\alpha\circ ev = \beta\circ ev$. Thus $ev: X\otimes X^\vee \to \mathbf{1}$ factors through the monomorphism ...

1

If $\mathcal{C}$ is a monoidal Krull-Schmidt category with finitely many indecomposables, then this translates to $r(\mathcal{C})$ being a free $\mathbb{Z}$-module of finite rank, so clearly an interesting example will be of this form. Take the ring to be $\mathbb{Z}[i]$ and note that if this arises as $r(\mathcal{C})$ for some category as above then this ...

2

For a symmetric (or even just braided) monoidal category, yes. Proof: \begin{align} \mathcal{C}(A,\text{Hom}(B,C)) & \cong \mathcal{C}(A\otimes B,C) \\ & \cong \mathcal{C}(B \otimes A, C)\text{ via braiding } \\ & \cong \mathcal{C}(B, \text{Hom}(A,C)) \end{align} So, $\text{Hom}(-,C)$ is adjoint to itself on the right, just as for cartesian ...

1

By no means. Think of the category of bimodules over a commutative ring $k$, so $k$-vector spaces with two actions which are associative with respect to each other but which may not agree. For instance $\mathbb{R}[\epsilon]$ has a natural extra action on all its modules by quotients go down to $\mathbb{R}$ and using the vector space action. Then $L(a,X)$ ...

2

You are more or less using the fact that $\text{Set}$ is symmetric monoidal with a very obvious associator, since its symmetric monoidal structure is the categorical product, together with the fact that the free vector space functor $\text{Set} \to \text{Vect}$ is symmetric monoidal. (The relevance of free vector spaces is that tensor products are ...

0

The construction you describe is exactly similar to the free-strict-monoidal category but it is not the free-monoidal category: the difference being that in a strict monoidal category associators and left and right units are required to be the identities of the respective objects, a requirement that you did not add in your construction. The ...

Top 50 recent answers are included