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It seems to me not, since this would seem to imply that for all functors F and all objects A in C there exists a morphism F(A) -> A (making all functors co-pointed?). However, intuitively it seems like the identity functor acts like a terminal object; a monad M on C is a monoid on [C, C] where the "unit" is a natural transformation η : I -> M, while for a monoidal set S in Set the unit is a function e : 1 -> S. So am I misunderstanding something, or are my intuitions leading me astray?

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The difference between those two examples is that in $\textbf{Set}$ the monoidal operation is the categorical product (so the identity object is the terminal object), whereas this is not true in the category of endofunctors on $\mathbf{C}$. (I believe the latter has a product if and only if $\mathbf{C}$ does, and then it is the pointwise product. It follows that the terminal object, if it exists, is the functor which sends all objects to $\mathbf{1}$ and all morphisms to the unique morphism $\mathbf{1} \to \mathbf{1}$. In particular, it's not the identity functor.)

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You mean "(so the identity FUNCTOR is...", don't you? –  a.r. Oct 8 '10 at 11:42
    
No, I mean the identity object. I'm comparing the monoidal structures on Set and on Fun(C, C). –  Qiaochu Yuan Oct 8 '10 at 12:57
    
"The difference between those two examples is that in Set the monoidal operation is the categorical product (so the identity object is the terminal object), whereas this is not true in the category of endofunctors on C." -- But the latter does have a monoidal product, which is functor composition. But I take it that "monoidal product" is not necessarily the same as "categorical product", even though in Set they happen to be the same thing (cartesian product). –  pelotom Oct 8 '10 at 18:30
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@pelotom: yes. A monoidal structure on a category doesn't have to be the categorical product; it can be the categorical coproduct or something weirder (for example the tensor product on vector spaces). –  Qiaochu Yuan Oct 8 '10 at 18:38

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