If $A$ is an algebra over $\mathbb C$, or, in other words, a monoid in the closed monoidal category $_{\mathbb C}\operatorname{Vect}$ of all vector spaces over $\mathbb C$, then, clearly, the category $_A\operatorname{Vect}$ of all left $A$-modules is enriched over the category $_{\mathbb C}\operatorname{Vect}$.

Is the same true if we replace $_{\mathbb C}\operatorname{Vect}$ by an arbitrary monoidal closed category $V$?


if we consider an arbitrary closed monoidal category $V$ and take an arbitrary monoid $A$ in $V$, will the category $_A V$ of all left $A$-modules be an enriched category over $V$?


You need that $V$ has equalizers (but for many basic stuff in the theory of $V$-enriched categories we need $V$ to be complete and cocomplete anyway). I denote internal Homs by $\underline{\mathrm{Hom}}$.

If $\underline{M} = (M, h : A \otimes M \to M)$, $\underline{N} = (N , k : A \otimes N \to N)$ are left $ A$-modules, define $\underline{\mathrm{Hom}}_A(\underline{M},\underline{N})$ as the equalizer of

$$\underline{\mathrm{Hom}}(M, N) \xrightarrow{h^*} \underline{\mathrm{Hom}}(M \otimes A,N)$$

and $$\underline{\mathrm{Hom}}(M,N) \xrightarrow{(\check{k})_*} \underline{\mathrm{Hom}}(M,\underline{\mathrm{Hom}}(A,N)) \xrightarrow{\cong} \underline{\mathrm{Hom}}(M \otimes A,N).$$

A reference for this is Proposition 1.2.17 in F. Marty, Des ouverts Zariski et des morphismes lisses en géométrie relative.

  • $\begingroup$ Martin, excuse me, for constructing this morphism $$ \operatorname{Hom}(M,N)\to \operatorname{Hom}(M,\operatorname{Hom}(A,N)), $$ I think we need the transformation of symmetry, $$ X\otimes Y\to Y\otimes X, $$ is that true? So your construction seems to work only for symmetric monoidal categories? (Or, I don't know, perhaps, all closed monoidal categories are assumed to be symmetric?) $\endgroup$ – Sergei Akbarov Apr 30 at 9:49
  • $\begingroup$ No, one uses (see my notation) the morphism $N \to \underline{\mathrm{Hom}}(A,N)$ which is adjoint to $A \otimes N \to N$. A symmetry is not necessary. $\endgroup$ – Martin Brandenburg Apr 30 at 15:28
  • $\begingroup$ (At least I think that it is not necessary. But I work with symmetric monoidal categories most of the time, so better doublecheck this.) $\endgroup$ – Martin Brandenburg Apr 30 at 15:50

Fact: an homomorphism of $A$-modules (for a monoid $A$ in a monoidal category $V$) is a map $f : M \to N$ such that $f(a.m)=a.fm$, or in purely diagrammatic form $f\circ \rho = \rho'\circ (A\otimes f)$, where $\rho : A \otimes M \to M$ and $\rho' : A\otimes N \to N$ are the actions that make $M,N$ two $A$-modules.

But then, the subobject of "homomorphisms of modules" is just the subobject of the hom-object $V(M,N)$ defined as the equaliser of two maps $$ \rho^* : V(M,N) \to V(A\otimes M,N) $$ and $$ \rho'_* \circ (A\otimes \_) : V(M,N) \to V(A\otimes M,A\otimes N) \to V(A\otimes M,N) $$

This definition gives you precisely what you want when the hom-objects are sets, but id "elementless".

Also, I'm using a general construction in a particular case: $A$-modules are algebras for the monad $T_A=A\otimes \_$, and $A$-module homomorphisms are $T_A$-algebra morphisms. If $T$ is any monad on a category $V$, the subobject of $T$-algebra morphisms between two $T$-algebras $(M, m : TM \to M)$ and $(N, n : TN \to N)$ is the equaliser of the pair of morphisms $$ m^* : V(M,N) \to V(TM,N) \qquad n_* \circ T_{MN} : V(M,N) \to V(TM,TN) \to V(TM, N) $$

  • $\begingroup$ Fosco, I don't understand this detail: $$ V(M,N)\to V(A\otimes M,A\otimes N). $$ To construct this morphism, we need a morphism $A\otimes M\to M$ and a morphism $N\to A\otimes N$, what are these two morphisms? Or I am missing something? $\endgroup$ – Sergei Akbarov Apr 30 at 7:40
  • $\begingroup$ @SergeiAkbarov No, one uses that $A \otimes -$ is a $V$-functor. $\endgroup$ – Martin Brandenburg Apr 30 at 8:08
  • $\begingroup$ @MartinBrandenburg: Martin, do you mean a functor between enriched categories? $\endgroup$ – Sergei Akbarov Apr 30 at 8:19
  • $\begingroup$ @SergeiAkbarov Yes. $\endgroup$ – Martin Brandenburg Apr 30 at 15:29
  • 1
    $\begingroup$ First lines of page 9: tac.mta.ca/tac/reprints/articles/10/tr10.pdf $\endgroup$ – fosco Apr 30 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.