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>A monad is a monoid in the category of endofunctors
Im not understanding this, as I understood it, its not.
a monoid is an object [math]X[/math] with 2 arrows [math]e: 1 \to X[/math] and [math]m: X \times X \to X[/math], where [math]X \times X[/math] is the product. The product of [math]T[/math]in the category of functors is [math]T \times T[/math] such that [math](T \times T)(X) = T(X) \times T(X)[/math], but for a monad the arrow is from [math]m: T(T(X)) \to T(X)[/math]. The only way this can make sense is if [math]T(X) \times T(X) = T(T(X))[/math] for all X. Is that true, or do I have an error somewhere? Or are they talking about 2 different products when they call the first one a product in the category of functors?
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>>9159689
How come four arrows in a square is math but one arrow pointing from A to B isn't?
>2occatl net /CG/ CG.pdf
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>>9159689
after some thought, I think I get it now, we can define a functor [math]\circ: T \times T \to T[/math] as [math]\circ: T \times T \to T\circ T[/math], then the multiplication for a monoid [math] m [/math] is related to the multiplication for a monad [math] \mu [/math] by [math]m_T = \mu(\circ(T\times T))[/math]
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Nah dude it's a burrito!
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>>9159689
yes, its a different notion of product.
Let End_C(X) be your category of endofunctors. The monoidal product is functor composition.
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>>9159917
yes
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