# reperiendi

## Monad for weakly monoidal categories

Posted in Category theory, Math by Mike Stay on 2009 August 19

We’ve got free and forgetful functors $L:\mbox{Cat} \to \mbox{WeakMonCat}, R:\mbox{WeakMonCat} \to \mbox{Cat}.$ Define $T = RL:\mbox{Cat} \to \mbox{Cat}.$ Given a category $X,$ the category $TX$ has

• binary trees with $\mbox{Ob}(X)-$labeled leaves as objects and
• binary trees with $\mbox{Mor}(X)-$labeled leaves together with the natural isomorphisms from the definition of a weakly monoidal category as its morphisms.

The multiplication $\mu_X:TTX\to TX$ collapses two layers of trees down to one. The unit $\eta_X:X \to TX$ gives a one-leaf tree.

An algebra of the monad is a category $X$ together with a functor $h:TX \to X$ such that $h \circ Th = h \circ \mu_X$ and $h \circ \eta_X = X.$ Define $x \otimes_X x' = h(\eta_X(x) \otimes_{TX} \eta_X(x')).$

Then the associator should be a morphism $a_X:(x \otimes_X x') \otimes_X x'' \to x \otimes_X (x' \otimes_X x'').$

However, it isn’t immediately evident that the associator that comes from $\otimes_{TX}$ does the job, since just applying $h$ to $a_{TX}$ gives $h((\eta_X(x) \otimes_{TX} \eta_X(x')) \otimes_{TX} \eta_X(x''))$

for the source instead of $h(\eta_X(h(\eta_X(x) \otimes_{TX} \eta_X(x'))) \otimes_{TX} \eta_X(x''))$,

which we get by replacing $\otimes_X$ with its definition above. We need an isomorphism $m:(x \otimes_X x') \otimes_X x'' \stackrel{\sim}{\to} h((\eta_X(x) \otimes_{TX} \eta_X(x')) \otimes_{TX} \eta_X(x''))$

so we can define $a_x = m^{-1} \circ h(a_{TX}) \circ m.$ Now we use the equations an algebra has to satisfy to derive this isomorphism. Since $h \circ Th = h \circ \mu_X,$ the following two objects are equal: $\begin{array}{rl} & h(\eta_X(h(\eta_X(x) \otimes_{TX} \eta_X(x'))) \otimes_{TX} \eta_X(x''))\\ = & h(\eta_X(h(\eta_X(x) \otimes_{TX} \eta_X(x'))) \otimes_{TX} \eta_X(h(\eta_X(x'')))) \\ = & (h \circ Th) ( \eta_{TX}(\eta_X(x) \otimes_{TX} \eta_X(x')) \otimes_{TTX} \eta_{TX}(\eta_X(x'')) \\ = & (h \circ \mu_X) ( \eta_{TX}(\eta_X(x) \otimes_{TX} \eta_X(x')) \otimes_{TTX} \eta_{TX}(\eta_X(x'')) \\ = & h((\eta_X(x) \otimes_{TX} \eta_X(x')) \otimes_{TX} \eta_X(x'')). \end{array}$

Therefore, the isomorphism $m$ we wanted is simply equality and $a_X = h(a_{TX}).$ It also means that $a_X$ satisfies the pentagon equation.

A similar derivation works for the unitors and the triangle equation.

A morphism of algebras is a functor $f:X \to Y$ such that $f \circ h_X = h_Y \circ Tf.$ Now $\begin{array}{rll} & f(x \otimes_X x') & \\ = & (f \circ h_X) (\eta_X(x) \otimes_{TX} \eta_X(x')) & \mbox{by de}\mbox{fn of }\otimes_X \\ = & (h_Y \circ Tf) (\eta_X(x) \otimes_{TX} \eta_X(x')) & \\ = & h_Y(\eta_X(f(x)) \otimes_{TX} \eta_X(f(x'))) & \mbox{by } T \\ = & f(x) \otimes_Y f(x')& \mbox{by de}\mbox{fn of }\otimes_Y\end{array}$

and $\begin{array}{rll} & f(I_X) \\ = & (f \circ h_X) (I_{TX}) & \mbox{by de}\mbox{fn of }I_X \\ = & (h_Y \circ Tf) (I_{TX}) & \\ = & h_Y(I_{TY}) & \mbox{since }T\mbox{ preserves the empty tree} \\ = & I_Y & \mbox{by de}\mbox{fn of }I_Y \end{array}$

so we have the coherence laws for a strict monoidal functor.

Also, $\begin{array}{rll} & f(a_X) & \\ = & (f \circ h_X) (a_{TX}) & \mbox{by the derivation above} \\ = & (h_Y \circ Tf) (a_{TX}) & \\ = & h_Y(a_{TY}) & \mbox{since }T\mbox{ preserves the associator} \\ = & a_Y & \mbox{again by the derivation above},\end{array}$

so it preserves the associator as well. The unitors follow in the same way, so morphisms of these algebras are strict monoidal functors that preserve the associator and unitors.

Advertisements