A question about monoidal functors

Question

For two monoidal categories $(M,\otimes)$ and $(N,\circledast)$, let $F:M \to N$ be a monoidal functor, and denote the structure natural isomorphism of $F$ by $J_{m,n}: F(m) \circledast F(n) \to F(m \otimes n)$. I guess it follows from the definition of coherence for a monoidal functor that we have, for any $m_i\in M$, a well-defined isomorphism $$ F(m_1) \circledast F(m_2) \circledast \cdots \circledast F(m_k) \to F(m_1 \otimes \cdots \otimes m_k)? $$ (Here for sake of simplicity I have supressed the associators.) Is this more or less immediate or is there some subtility in this?

Answer 1

nah, that is it, also, you dont need any associator at that point, both tensorproducts are implicitly considered bracketed from right to left (standard bracketing for any consecutive operation), so that all rolls through smoothly (composition of literally $k-1$ many natural isomorphisms $J_{m_{i+1},m_{i}}$).

More precisely: $$J_{m_1,...,m_k}:=J_{m_1,\left(m_2\otimes...\otimes m_k\right)}\circ J_{m_2,\left(m_3\otimes...\otimes m_k\right)}\circ...\circ J_{m_{k-1},m_k}:F\left(m_1\right)*...*F\left(m_k\right)\to F\left(m_1\otimes...\otimes m_k\right)$$

Sorry, did not know how to do your fancy star. Hope that clarifies things though.

Last funfact: Observe that the above construction is also the natural way to compose the above natural isomorphisms (sending them the other way around via their inverses would make trouble with compositions), which explains why the idea of a lax monoidal functors are the natural generalization.

Answer 2

For simplicity, and to highlight the subtlety, let me assume we have strict monoidal categories. A lax monoidal functor $F$ is equipped with morphism $$\eta : e \to F (e)$$ and a natural transformation $$\mu_{x, y} : F (x) \otimes F (y) \to F (x \otimes y)$$ that make the following diagrams commute: $$\require{AMScd} \begin{CD} F (x) \otimes F (y) \otimes F (z) @>{\mu_{x, y} \otimes \textrm{id}_{F (z)}}>> F (x \otimes y) \otimes F (z) \\ @V{\textrm{id}_{F (x)} \otimes \mu_{y, z}}VV @VV{\mu_{x \otimes y, z}}V \\ F (x) \otimes F (y \otimes z) @>>{\mu_{x, y \otimes z}}> F (x \otimes y \otimes z) \end{CD}$$ $$\begin{CD} e \otimes F (x) @>{\eta \otimes \textrm{id}_{F (x)}}>> F (e) \otimes F (x) \\ @| @VV{\mu_{e, x}}V \\ F (x) @= F (e \otimes x) \end{CD}$$ $$\begin{CD} F (x) \otimes e @>{\textrm{id}_{F (x)} \otimes \eta}>> F (x) \otimes F (e) \\ @| @VV{\mu_{x, e}}V \\ F (x) @= F (x \otimes e) \end{CD}$$ From the above three axioms, we can deduce that "all diagrams commute". I will not attempt to state this precisely; instead, let me give an example. Suppose we seek a natural transformation: $$F (w) \otimes F (x) \otimes F (y) \otimes F (z) \to F (w \otimes x \otimes y \otimes z)$$ One obvious candidate is: $$\mu_{w \otimes x \otimes y, z} \circ (\mu_{w \otimes x, y} \otimes \textrm{id}_{F (z)}) \circ (\mu_{w, x} \otimes \textrm{id}_{F (y) \otimes F (z)})$$ But we also have its mirror: $$\mu_{w, x \otimes y \otimes z} \circ (\textrm{id}_{F (w)} \otimes \mu_{x, y \otimes z}) \circ (\textrm{id}_{F (w) \otimes F (x)} \otimes \mu_{y, z})$$ Or we could take a balanced approach: $$\mu_{w \otimes x, y \otimes z} \circ (\mu_{w, x} \otimes \mu_{y, z})$$ You could even use $\eta$ here and there if you like. The axioms say that these formulas all name the same morphism. Indeed, we have $$\mu_{w \otimes x \otimes y, z} \circ (\mu_{w \otimes x, y} \otimes \textrm{id}_{F (z)}) \circ (\mu_{w, x} \otimes \textrm{id}_{F (y) \otimes F (z)}) = \mu_{w \otimes x, y \otimes z} \circ (\textrm{id}_{F (w \otimes x)} \otimes \mu_{y, z}) \circ (\mu_{w, x} \otimes \textrm{id}_{F (y) \otimes F (z)})$$ and $$(\textrm{id}_{F (w \otimes x)} \otimes \mu_{y, z}) \circ (\mu_{w, x} \otimes \textrm{id}_{F (y) \otimes F (z)}) = \mu_{w, x} \otimes \mu_{y, z}$$ so indeed: $$\mu_{w \otimes x \otimes y, z} \circ (\mu_{w \otimes x, y} \otimes \textrm{id}_{F (z)}) \circ (\mu_{w, x} \otimes \textrm{id}_{F (y) \otimes F (z)}) = \mu_{w \otimes x, y \otimes z} \circ (\mu_{w, x} \otimes \mu_{y, z})$$ Symmetrically, $$\mu_{w, x \otimes y \otimes z} \circ (\textrm{id}_{F (w)} \otimes \mu_{x, y \otimes z}) \circ (\textrm{id}_{F (w) \otimes F (x)} \otimes \mu_{y, z}) = \mu_{w \otimes x, y \otimes z} \circ (\mu_{w, x} \otimes \mu_{y, z})$$ so the three candidates above are indeed equal.

---

For completeness, let me give an example of a functor with the data $\eta$ and $\mu$ but which does not satisfy the axioms.

Let $T$ be any magma and let $u$ be any element of $T$. Let $\mathbf{S}$ be (a strictified version of) the cartesian monoidal category of sets. We have a functor $F : \mathbf{1} \to \mathbf{S}$ sending the unique object $\bullet$ of $\mathbf{1}$ to $T$. Let $\eta : 1 \to F (\bullet)$ be the map sending the unique element of $1$ to $u$, and let $\mu_{\bullet, \bullet} : F (\bullet) \times F (\bullet) \to F ({\bullet} \otimes {\bullet})$ be the given binary operation $T \times T \to T$. Then $F$, $\eta$, and $\mu$ constitute the data of a lax monoidal functor if and only if $T$ is a monoid with the given binary operation and $u$ is a unit for that binary operation.

If you want an example where $\eta$ and $\mu$ are isomorphisms, replace $\mathbf{S}$ with the category of vector spaces. A bilinear operation on a 1-dimensional vector space is automatically unital and associative, but not every element is a unit, so the axioms involving $\eta$ can fail in this case.