monoidality of $A\otimes (-)$ with $A$ monoid belonging to the center

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

• Is 'ecc.' meant to be 'etc.'? Jul 27, 2019 at 19:45
• 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. Jul 27, 2019 at 21:44
• I'd say go ahead and post details (it would save me from having to write it up myself). Jul 27, 2019 at 21:48
• does 'monoidal' mean 'lax monoidal'? Aug 25, 2020 at 11:43

1 Answer

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:

$$AXAYAZ\to AXAAYZ\to AXAYZ\to AAXYZ\to AXYZ$$

proceed:

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

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.