I am trying to generalize the notion of monoid object internal to a (not necessarily strict) monoidal category, by weakening the associativity and unitarity diagrams (see this nlab entry.) Of course this construction only makes sense in a (not necessarily strict) monoidal 2-category, since we use 2-isomorphisms. As written in the nlab entry, the goal is to be able to define a (not necessarily strict) monoidal category as a pseudomonoid in $\textbf{Cat}$. This is my definition so far:

Given a (not necessarily strict) monoidal 2-category $(C,\otimes, I, \alpha_C, \lambda_C, \rho_C)$, where $\alpha_C$, $\lambda_C$ and $\rho_C$ are respectively the associator and the left and right unitors, we define a pseudomonoid in $C$ by the following data:

  • an object $M \in |C|$
  • a 1-morphism $\mu: M \otimes M \to M$ called multiplication;
  • a 1-morphism $\eta: I \to M$ called the unit;
  • a 2-isomorphism $\alpha_M: \mu \circ (\mu \otimes \text{Id}_M) \to \mu \circ (\text{Id}_M \otimes \mu) \circ \alpha_C$ called the (internal) associator;
  • a 2-isomorphism $\lambda_M: \mu \circ (\eta \otimes \text{Id}_M) \to \lambda_C$ called the (internal) left unitor, and
  • a 2-isomorphism $\rho_M: \mu \circ (\text{Id}_M \otimes \eta) \to \rho_C$ called the (internal) right unitor,

such that some diagrams commute.

As you might expect I am having some trouble defining the relevant diagrams, namely MacLane's pentagon and the triangle identity. Normally $\alpha$ is defined my its morphisms $\alpha_{x,y,z}: x \cdot (y \cdot z) \to (x \cdot y) \cdot z$ where $a \cdot b$ is shorthand for $\mu(a,b)$. That gives a very nice way of writing the pentagon. Simlarly, $\lambda$ is defined by $\lambda_x: 1 \cdot x \to x$ and $\rho$ by $\rho_x: x \cdot 1 \to x$, where $1$ is shorthand for $\eta(I)$ I suppose.

However, defining $\alpha$ this way seems evil to me (not in the technical sense), since $M$ need not contain elements, and in general 2-morphisms are not natural transformations. In the spirit of category theory, is there a way to define the 2-isomorphisms (maybe like I did above?) in a way that makes the relevant diagrams possible to write?

As might be apparent, I am not very familiar with this kind of 2-categorical reasoning, and as far as I know this question may not even be well-defined or even meaningful. I apologize in advance if that is the case.

  • $\begingroup$ Don't be so modest, this is a very well written question. $\endgroup$ Jun 10, 2014 at 13:11
  • $\begingroup$ Thank you, your comment makes me feel less out of place. $\endgroup$
    – JoeyBF
    Jun 10, 2014 at 13:23

1 Answer 1


The pentagon identity is a commutative diagram of $2$-morphisms, each one induced by $\alpha_M$. It looks as follows. Because I'm lazy, I will omit the associators from $\mathcal{C}$.

$$\begin{array}{cc} \mu \circ ((\mu \circ (\mu \otimes\mathrm{id}_M)) \otimes \mathrm{id}_M) & \rightarrow & \mu \circ ((\mu \circ (\mathrm{id}_M \otimes \mu)) \otimes \mathrm{id}_M) & \rightarrow & \mu \circ (\mathrm{id}_M \otimes (\mu \circ (\mu \otimes \mathrm{id}_M))) \\ \downarrow &&&& \downarrow \\ \mu \circ (\mu \otimes \mu) & & \rightarrow && \mu \circ (\mathrm{id}_M \otimes (\mu \circ (\mathrm{id}_M \otimes \mu))) \end{array}$$ It should be rather straight forward to write down the $2$-morphisms here. For example, the first $\rightarrow$ in the first line is $\mu \circ (\alpha_M \otimes \mathrm{id}_M)$.

Applying this to the monoidal $2$-category of small categories, we obtain the usual pentagon identity.

$$\begin{array}{c} ((A \otimes B) \otimes C) \otimes D & \rightarrow & (A \otimes (B \otimes C)) \otimes D & \rightarrow & A \otimes ((B \otimes C) \otimes D) \\ \downarrow &&&& \downarrow \\ (A \otimes B) \otimes (C \otimes D) && \rightarrow && A \otimes (B \otimes (C \otimes D))\end{array}$$

  • $\begingroup$ Thank you for your answer. If I may, I have one more question. If you added the associators from $C$, wouldn't the cells of your diagrams become 2-morphisms? I would see it as whiskering $\alpha_C$ by what you wrote, and then the morphisms between these would be 3-morphisms. Where have I made a mistake? $\endgroup$
    – JoeyBF
    Jun 10, 2014 at 15:23
  • $\begingroup$ No, $\alpha$ is a $2$-morphism. $\endgroup$ Jun 10, 2014 at 17:14

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