Consider category of vector spaces. Consider functors from it to itself. They actually form an algebra - since vector spaces can be added and tensor multiplied.

Question Is there co-product on this algebra ? If yes, how to construct/motivate/... it ? If yes can it be explained from categorical point of view ? Can it be generalized to other categories ?

I guess the answer is YES. The reason is the following - as far as I understand algebra of endo-functors of Vect is isomorphic to the algebra of symmetric functions.

[EDIT] this is wrong (Thanks to Martin). Automorphism $C$ of any object $V$ gives end-functor twisting morphisms by $C$ and/or $C^{-1}$ ( i.e. $Mor(V,A)$ twisted by $C$ , $Mor(A,V)$ twisted by $C^{-1}$, hence $Mor(V,V)$ twisted by $C ...C^{-1}$).

So Schur functors are NOT all end-functors, so probably the question should be revised, how to characterize Schur functors among all functors and how to define co-product for them. [End EDIT]

See e.g. third paragraph in Qiaochu Yuan answer here:

Categorification and Schur functors

Remark: Schur functors corresponds to Schur functions.


On the other hand there is co-multiplication on the algebra of symmetric functions

Symmetric polynoms are Hopf algebra ? What for one needs co-product ?

So since two algebras are isomorphic means both of them are co-algebras.

In the question about co-product on symmetric function algebra I got a lots of beautiful answers. However to my taste all of them somewhat tricks - it is not so clear (for me) why this cosntruction is somewhat natural. So may be categorical point of view may clarify.



Operations of functors are defined as follows: take F,G: Vect-> Vect we want to multiply them, i.e. to define new functor $FG$, it is defined on objects as follows:

$FG: V-> F(V)\otimes G(V)$,

and on morphisms it is defined respectively $\phi: V->W$,

$FG(\phi): F(V)\otimes G(V) \to F(W)\otimes G(W)$

$FG(\phi)= F(\phi) \otimes G(\phi)$.


absolutely the same with summation - we should substitute tensor product by the direct sum.


Exercise: functors form an associative algebra with repsect to these operations.

-- [EDIT]


Answering comments by Martin and Qiaochu. I would prefer to call by "Vect" what they (and may be everybody) call "skeleton of Vect". i.e. it is category where for each natural n there is only one object, and tensor product C^nC^m = C^nm just equal without any isomorphisms. This well-defined category and let me work with it.

  • $\begingroup$ You probably mean the algebra of isomorphism classes of endofunctors of fdVect. Otherwise you get a 2-algebra. $\endgroup$ Feb 26, 2012 at 10:46
  • $\begingroup$ @Martin, may be I am wrong, but I do not understand what is the problem: if I have two functors I can add and multiply them - so get a ring . Is it Okay ? $\endgroup$ Feb 26, 2012 at 11:00
  • $\begingroup$ @Alexander: the problem is that the tensor product is not strictly associative; it's only associative up to a canonical isomorphism, the associator. To get something associative on the nose you need to work with isomorphism classes. $\endgroup$ Feb 26, 2012 at 11:19
  • $\begingroup$ By the way there are lots of endofunctors of fdVect which are not Schur functors. $\endgroup$ Feb 26, 2012 at 11:23
  • $\begingroup$ @Qiaochu what does it mean? For Vect I do not quite see problems. @Martin Schur functions are linear basis, but linear combination is not schur functor $\endgroup$ Feb 26, 2012 at 11:43

1 Answer 1


Let ${\bf V}$ be the category of endofunctors (or polynomial endofunctors or whatever condition you want). A comultiplication correspond to a functor ${\bf V} \to {\bf V} \times {\bf V}$ satisfying some properties. One way to construct this is to compose a functor $Vec \to Vec$ with a functor of the form $Vec \times Vec \to Vec$. There are two natural candidates: $(A, B) \mapsto A \oplus B$ and $(A, B) \mapsto A \otimes B$.

The first one is the one that corresponds to the usual comultiplication on the ring of symmetric functions, i.e., the one defined by $f \mapsto f(x,y)$ where $x,y$ are distinct alphabets.


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