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}


\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.


\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.


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