Let $(\mathcal{C}, \otimes)$ a monoidal category, and $(A, m, e)$ a monoid (where $m: A\otimes A\to A$, $e: I\to A$ ecc. ), with $(A, u)$ belonging to the centre of $(\mathcal{C}, \otimes)$: $u: A\otimes (-)\cong (-)\otimes A$, ecc. see [JS] p.38.

Consider tha (usual) functor $F_A(-):= A\otimes (-): \mathscr{C}\to \mathscr{C}$, there are the natural morphisms:

$\alpha_{X,Y}: F_A(X)\otimes F_A(Y)= A\otimes X\otimes A\otimes Y\xrightarrow{1u1} A\otimes A\otimes X\otimes Y \xrightarrow{m11} A\otimes X\otimes Y=F_A(X\otimes Y)$

$\phi: I\cong I\otimes I\xrightarrow{e1}A\otimes I=F_A(I)$.

Question: Is the data above define a monoidal functor? What coherence axioms (between the monoidal and centre object structure) we need?

[JS] Braided Tensor Categories, A.Joyal, R.Street https://www.sciencedirect.com/science/article/pii/S0001870883710558

  • 1
    $\begingroup$ Is 'ecc.' meant to be 'etc.'? $\endgroup$
    – LSpice
    Jul 27, 2019 at 19:45
  • $\begingroup$ Yes, I'm Italian, my English is very imperfect. Anyway I think to resolved my question: the axiom consist in the equality of the following two compositions: $A\otimes X\otimes A\otimes Y\to X\otimes A\otimes A\otimes Y\to X\otimes A\otimes Y\to A\otimes X\otimes Y$ and $A\otimes X\otimes A\otimes Y\to A\otimes A\otimes X\otimes Y\to A\otimes X\otimes Y$, if you find interesting I'll post details. $\endgroup$ Jul 27, 2019 at 21:44
  • 3
    $\begingroup$ I'd say go ahead and post details (it would save me from having to write it up myself). $\endgroup$
    – Todd Trimble
    Jul 27, 2019 at 21:48
  • $\begingroup$ does 'monoidal' mean 'lax monoidal'? $\endgroup$ Aug 25, 2020 at 11:43

1 Answer 1


I write $XY$ instead of $X\otimes Y$ and ignore canonical isomorphism and morphisms labels (unambiguous).

I assume the axiom:

Ax) $(AXAY\to AAXY\to AXY) = (AXAY\to XAAY\to XAY\to AXY)$.

The second member is an alternative computation of $\alpha$.

First I have to show that:

$F_A(X)F_A(Y)F_A(Z)\to F_A(XY)F_A(Z)\to F_A(XYZ) =$

$F_A(X)F_A(Y)F_A(Z)\to F_A(X)F_A(YZ)\to F_A(XYZ)$

i.e. that $$AXAYAZ\to AAXYAZ\to AXYAZ\to AXAYZ\to AAXYZ\to AXYZ$$

is equal to:



$(AXAYAZ\to AAXYAZ\to AXYAZ)\to (AXAYZ\to AAXYZ\to AXYZ)=^{Ax} =(AXAYAZ\to XAAYAZ\to XAYAZ\to AXYAZ) \to (AXAYZ\to XAAYZ\to XAYZ\to AXYZ)=^{natuality}$ $=AXAYAZ\to XAAYAZ\to (XAYAZ \to XAAYZ\to XAYZ\to AXYZ)=^{Ax}$ $=AXAYAZ\to XAAYAZ\to (XAYAZ \to XYAAZ\to XYAZ\to XAYZ\to AXYZ)=^{Ax}$ $=A_1XA_2YA_3Z\to XA_1A_2YA_3Z\to XA_{1,2}YA_3Z \to XA_{1,2}A_3YZ\to XAYZ\to AXYZ=^{naturality}$ $=A_1XA_2YA_3Z\to XA_1A_2YA_3Z\to XA_1A_2A_3YZ\to XA_{1,2}A_3YZ \to XAYZ\to AXYZ=^{monoid}$ $=A_1XA_2YA_3Z\to XA_1A_2YA_3Z \to XA_1A_2A_3YZ\to XA_1A_{2,3}YZ \to XAYZ\to AXYZ=^{naturality}$ $=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to XA_1A_2A_3YZ \to XA_1A_{2,3}YZ \to XAYZ\to AXYZ=^{naturality}$ $=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to (A_1XA_{2,3}YZ\to XA_1A_{2,3}YZ \to XAYZ\to AXYZ)=^{Ax}$ $=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to A_1XA_{2,3}YZ\to A_1A_{2,3}XYZ \to AXYZ.$

About the unity axioms we prove that $F_A(X)\cong IF_A(X)\to F_A(I)F_A(X)\to F(IX)\cong F(X)$ is the identity, this is::

$AX\cong IAX\to AAX\cong AIAX\cong IAAX\to IAX\cong AX$ for naturality this is:

$AX\cong IAX\to AAX\cong AIAX\cong IAAX\cong AAX\to AX$ for the usual monoidal topic this is :

$AX\cong IAX\to AAX\to AX=1$. The other unitary axiom is quite similar.


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