6
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ Is 'ecc.' meant to be 'etc.'? $\endgroup$ – LSpice Jul 27 '19 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. $\endgroup$ – Buschi Sergio Jul 27 '19 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 '19 at 21:48
  • $\begingroup$ does 'monoidal' mean 'lax monoidal'? $\endgroup$ – Dylan Wilson Aug 25 '20 at 11:43
5
$\begingroup$

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.

$\endgroup$

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.