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

Cassiopeia’s veil

Posted in Uncategorized by Mike Stay on 2009 August 13

IMG_1577

Around work

Posted in Uncategorized by Mike Stay on 2009 August 13