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Stream: learning: questions

Topic: how to argue quotient maps form a natural transformation?

Naso (May 28 2023 at 04:41):

Let $F : \mathcal{Set} \to \mathcal{Semi}$ be the free semigroup functor.

For any set $A$ there is a semigroup $I(A)$ called the 'free semigroup with idempotent generators', which is $F(A)/\sim$ where $\sim$ is the smallest congruence containing the pairs $(a,aa)$ for each $a \in A$.

Now for each $A$ we have a quotient map $q_A : F(A) \to I(A)$.

Let $\Omega : \mathcal{T}^\mathsf{op} \to \mathcal{Set}$ be a presheaf on a topological space $(X, \mathcal{T})$.

I want to argue that these quotient maps assemble to a natural transformation $q : F \circ \Omega \to I \circ \Omega$.

"It seems obvious" (famous last words) and yet I can't think of a concise way to explain it.

John Baez (May 28 2023 at 04:58):

I think that the maps $q_A : F(A) \to I(A)$ are components of a natural transformation $q: F \Rightarrow I$. (I write natural transformations with a double arrow.) You can then right whisker this natural transformation with the functor $\Omega$ to get a natural transformation $q \circ \Omega: F \circ \Omega \Rightarrow I\circ \Omega$. You are calling this latter natural transformation $q$.