I'm finally learning homological algebra, and the notion of an enriched category seems to be the right setting in which to define the $\mathbf{Ext}$ functor. Yet, the definitions of an enriched category that I've seen require the $\operatorname{Hom}$ objects to be objects of a monoidal category, so that one is able to define composition as a morphism $$ \operatorname{Hom}(B,C) \otimes \operatorname{Hom}(A,B) \to \operatorname{Hom}(A,C), $$ the intuition being that composition is bilinear, so that the universal property of the tensor product may be applied.

Yet, is it not sufficient to require the existence of products and linearity (or whatever property is required in order to have a morphism) in each component when the other component is left fixed?

  • $\begingroup$ What's the difference between "bilinear" and "linear in each component"? $\endgroup$ Feb 26, 2019 at 11:11
  • $\begingroup$ There isn't any. I clarified. $\endgroup$ Feb 26, 2019 at 11:11
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    $\begingroup$ It is still not clear to me what you mean. The object $\mathrm{Hom}(B, C)$ is (in this case) an object of some category -- you cannot a priori assume any internal structure of it. In particular it doesn't necessarily make sense to think of it as consisting of points (morphisms) which can have some structure. $\endgroup$ Feb 26, 2019 at 11:43
  • $\begingroup$ What does "the other component is fixed" mean ? In this context, $\hom (A,B)$ is not necessarily a set, or doesn't even need to have an underlying set, so what would that mean ? $\endgroup$ Feb 26, 2019 at 11:45
  • $\begingroup$ @Max Either we would have a locally small category, or $\operatorname{Hom}(A,B)$ would be a proper class with algebraic structure such as a Field, where the capital initial letter indicates that its elements form a proper class (an example of such an object would be the surreal numbers). "The other component is fixed" means for example that whenever $g: B \to C$ is a morphism, then composition $g \circ \cdot$ is a morphism in whatever category one has enriched oneself. $\endgroup$ Feb 26, 2019 at 12:10

1 Answer 1


I am not exactly sure what your question means.

If I have a category $\mathcal{C}$, that I want to be enriched over a category $D$, but I don't suppose a priori that $\mathcal{D}$ has a monoidal structure.

Now to define the composition as a map in $\mathcal{D}$, what do I need?

  • First, I have to take two composable $Hom$ objects : $Hom(A,B)$ and $Hom(B,C)$, and I want a map that ends up in $Hom(A,C)$. I don't know yet what should be the source of this map, but I know that it has to be computed inside $\mathcal{D}$ from the two existing objects $Hom(A,B)$ and $Hom(B,C)$. So for now I will denote $Hom(B,C)\boxtimes Hom(A,B)$ this object, so that the composition of morphisms is actually a map $\circ: Hom(B,C)\boxtimes Hom(A,B) \to Hom(A,C)$ in $\mathcal{D}$. I will try to make as little assumptions as possible on the operation $\boxtimes$, while still making is a composition.

  • I want this operation to be functorial, so that I can use $circ : Hom(B,C)\boxtimes Hom(A,B) \to Hom(A,C)$ to define an arrow $Hom(C,D)\boxtimes(Hom(B,C)\boxtimes Hom(A,B)) \to Hom(C,D) \boxtimes Hom(A,C)$. In general this means that my composition will respect the other computations that can happen inside $\mathcal{D}$, which is the whole point of enriched categories.

  • I know that I want the composition to be assocative, in other words, I want the two compositions $Hom(C,D)\boxtimes (Hom(B,C) \boxtimes Hom(A,B)) \to Hom(C,D)\boxtimes Hom(A,C) \to Hom(A,D)$ $(Hom(C,D)\boxtimes Hom(B,C))\boxtimes Hom(A,B) \to Hom(B,D)\boxtimes Hom(A,B) \to Hom(A,D)$ to be equal. But this doesn't a priori makes sense because their sources are different, so these two arrows are not comparable. the only way to make these arrows comparable is to assume that there is an invertible morphism $\alpha : Hom(C,D)\boxtimes (Hom(B,C) \boxtimes Hom(A,B)) \simeq (Hom(C,D)\boxtimes Hom(B,C)) \boxtimes Hom(A,B)$ in $\mathcal{D}$

  • Next I want each of the Hom(A,A) to have an identity. Given that my $Hom$ are not sets anymore, I can't talk about their elements, but I can use the trick of saying that an element is the same of a map from a distinguished object. I have to have a distinguished object $I$ in $\mathcal{D}$. Even by being as general as possible, and assuming nothing yet on $I$, I have a map $i : I \to Hom(A,A)$ in $\mathcal{D}$.

  • Now I also want the identity to be a left unit, that is considering $I\boxtimes Hom(A,B) \to Hom(B,B) \boxtimes Hom(A,B) \to Hom(A,B)$ should be the identity of $Hom(A,B)$. But again this is not possible, since its source is $I\boxtimes Hom(A,B)$, and not just $Hom(A,B)$. So to express this property, I need to have a morphism $I\boxtimes Hom(A,B) \to Hom(A,B)$ in $\mathcal{D}$.

You could do the same thing for the right unit, and you finally get that $\boxtimes$ has to satisfy all the axioms of a monoidal structure on the category $\mathcal{D}$. So in the end, if you want to enrich the category $\mathcal{C}$ over the category $\mathcal{D}$, in order to express all the properties you want the composition to satisfy, you need $\mathcal{D}$ to be a monoidal category, and so this really is the context in which you want to work.

I am not really sure what you call linearity in this context, given that there is a priori no addition, nor scalar multiplication of the objects of the category $\mathcal{D}$, and monoidal categories have a definition more general than the tensor product on the category of modules.

  • $\begingroup$ As I explained in the comments, I was attempting to generalise the notion of a category in a different way to a similar concept. The point would have been that on the one hand, abelian categories become categories that are enriched over abelian groups, and on the other hand, one might have dropped certain conditions on the category one is enriched over. But the fact that it seems as though one may deduce all axioms from the basic requirements convinces me that the standard notion of an enriched category is the correct one (I'll still have to work out the pentagonal identity). $\endgroup$ Feb 26, 2019 at 12:40
  • $\begingroup$ I'm just realising that I'll have to learn category theory all over again, thanks to this more general setting. (Now I finally understand why Grothendieck in his Tohoku paper defines every limit with respect to $\operatorname{Hom}$ classes.) $\endgroup$ Feb 26, 2019 at 12:52

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