# Monoidal functor and the units II

In their book Tensor Categories Etingof, Gelaki, Nikshych and Ostrik give a different definition of a (strong) monoidal functor. The difference is that they do not set the isomorphism $$F(1) \cong 1$$ as a part of the data, but rather impose a condition on the pair $$(F\colon C\to C', (J_{X,Y}\colon F(X)\otimes F(Y)\to F(X\otimes Y))_{X,Y \in C})$$ that would be a monoidal functor that some isomorphism $$F(1) \cong 1$$ exists. They then define a canonical isomorphism $$F(1)\cong 1$$ by the following diagram:

The exercise is then given to prove that, for this canonical isomorphism, the following diagrams commute:

My trouble is the second diagram. Clearly, we first should tensor diagram defining $$\phi$$ with $$F(X)$$ and $$1_{F(X)}$$ and appeal to functoriality, but I don't see how we can get from $$1_{F(X)}\otimes (\phi \otimes 1_1)$$ to $$1_{F(X)}\otimes \phi$$.

• Can't you just take the diagram defining $\varphi$ and apply $-\otimes F(X)$ to it ? Then you draw another square which is the same but the $X$ is inside the rightmost $F$ (e.g. the top right hand corner is $F(1\otimes X)$) and relate those two squares via the appropriate $J$, then use the unitors for $C$ ? Commented Dec 9, 2019 at 20:31
• I prefer to think in terms of diagrams than equations (it's way easier for me at least), but I just wrote it down, and I think the second digram should have $l_{F(1\otimes X)}$ as top map, $F(l_{1\otimes X}^{-1})$ as rightmost map, $\varphi\otimes id_{F(1\otimes X)}$ as leftmost map and $J_{1,1\otimes X}$ as bottom map. Commented Dec 9, 2019 at 21:43
• You don't know that it does yet; but the point is that you can prove it is isomorphic to the first square, which does. Therefore this one does too ! And then you can relate it to the desired square via $F$ of various unitors Commented Dec 9, 2019 at 22:27
• Monoidal functors play well with associators, so there should be no problem there Commented Dec 13, 2019 at 11:14

As Max already pointed out, you may start with tensoring the diagram (2.24) defining $$\varphi$$ with $$F(X)$$. So you get the commutative diagram (1): Notice, that (1) looks like the desired diagram (2.26), tensored with $$F(1)$$: the diagram (2): Now we are going to show that (2) is isomorphic to (1), which immediately implies that (2.26) is commutative. To prove that (2) is isomorphic to (1) we should prove the commutativity of the certain cube with (1) and (2) as opposite faces. To indicate faces I will use the Rubik's Cube notation. Define the B-side as (1) and F-side as (2). Set the U-side as the following diagram:
which is commutative by the triangle diagram. The L-side is simply associativity, so I will not draw it. The D-side is following: It is commutative by the monoidal structure axiom. The R-side is automatically defined, let's prove its commutativity. Indeed, $$J_{X\otimes 1,1}^{-1}\circ F(a_{X,1,1}^{-1})\circ J_{X,1\otimes 1}\circ(\text{id}_{F(X)}\otimes^{\wr}F(\ell_1^{-1}))= J_{X\otimes 1,1}^{-1}\circ F(a_{X,1,1}^{-1})\circ F(\text{id}_X\otimes\ell_1^{-1})\circ J_{X,1}=$$ $$J_{X\otimes 1,1}^{-1}\circ F(a_{X,1,1}^{-1}\circ(\text{id}_X\otimes\ell_1^{-1}))\circ J_{X,1}= J_{X\otimes 1,1}^{-1}\circ F(r^{-1}_X\otimes\text{id}_1)\circ J_{X,1}= F(r^{-1}_X)\otimes^{\wr}\text{id}_{F(1)},$$ where the first and fourth equalities are the naturality of $$J$$, the second is the functoriality of $$F$$ and the third is the triangle diagram for the domain monoidal category.
• Astonishingly done, Oskar! One last thing, though, is to relate (2.26) tensored by $F(1)$ (your (2)) to (2.26) tensored by $1$ via $\phi$ in an another (but much simplier) cube, and prove that the latter commutes. Then we can use the faithfulness of action of $-\otimes 1$ on a category, and obtain the commutativty of (2.26). Am I correct? Commented Dec 16, 2019 at 21:17
• @Jxt921 You are welcome. In fact, $(2.26)\otimes F(1)$ is isomorphic to $(2.26)\otimes1$ (it is quite obvious) and $(2.26)\otimes 1$ is isomorphic to $(2.26)$ (because $r$ is a natural iso, or by faithfulness of $-\otimes1$), therefore all these diagrams are equicommutative (hence commutative). Commented Dec 16, 2019 at 21:37