# Create 2-category from a cartesian category

I have a way of constructing (something like) a 2-category from a category with products $\mathcal{C}$.

My question: Is this a correct construction, and if so, does it have a name?

Define $\mathcal{C'}$ to consist of one object. The 1-morphims are objects of $\mathcal{C}$, the identity is the terminal object of $\mathcal{C}$, and $A \circ B$ is $A \times B$. The 2-morphisms $A \to B$ are morphisms $A \to B$ from $\mathcal{C}$. Vertical composition is composition in $\mathcal{C}$ and the horizontal composition of $f$ and $g$ is given by $f \times g$.

Notes:

1. I said "something like" a 2-category because it seems like composition will be defined up to isomorphism, rather than equality.

2. Perhaps this would work with any monoidal category, rather than just a category with products.

3. I want this construction so I can "reverse" the usual relationship between monoid object and monad, and say that a monoid in $\mathcal{C}$ is given by a monad in $\mathcal{C'}$.

• Take a look at the bottom of page three here: math.berkeley.edu/~aaron/livetex/highercats.pdf Also the bottom of page 5 is illuminating. Oct 20 '13 at 2:26
• This is a correct construction (though by "terminal" object, you probably mean unit object) and should work for any monoidal structure. I'm not sure if there's a name for it, but higher algebra people think about this kind of construction often. It shows you can trade off between category number and number of compatible multiplications (composition and monoidal structure give 2 multiplications). For instance, an (oo,n)-category with only n-morphisms is the same thing as an E_n algebra. Oct 20 '13 at 3:00
• @BabyDragon: Ah it is a correspondance in both directions! That makes sense. I didn't even consider that. Oct 20 '13 at 11:40