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6272150 No.6272150 [Reply] [Original]

Can someone elaborate on the Yonneda lemma in the case where the functor is a sheaf?
-Why- are the natural transformations (in bijection with) the image of the functor? Because the image are usually not a priori a set of functions, right?, so this is indeed surprising to me. I read answers in terms of something is generated, but I'm not smarter.

>> No.6272203

The image can be anything. The lemma just states that there is a bijection between F(A) and the natural tranformations between F and the hom functor A. I don't see how F being a sheaf changes anything.

>> No.6272205

>>6272203
>The image can be anything.
Except you can clearly see that it is a bird moving left while the caption "What are birds?" is displayed.

>> No.6272207

>>6272205
lel

>> No.6272206

>>6272203
By anything I mean of course any set.

>> No.6272212

>>6272203
I think he's asking for intuition in the case F is a sheaf.

>> No.6272228

>>6272212
that's right. When i think of sheaves I want to think of geometrical theories, bundles etc. And I don't see how the relation between the image of the hom functor and these geometrical objects is given by "the same" objects as the set which is the image - some geometrical thing.

>> No.6272314

>>6272228
Unfortunately I won't be able to help there. I'm learning this stuff without much more than introductory algebraic topology. A sheaf to me is just an Functor satisfying certain properties.

>> No.6272413

In short: the Yoneda lemma doesn't tell you anything geometrically interesting for sheaves. All it really says is that presheaves in some precise sense generalize your open sets. Sheaves are then a controlled and more well-behaved version of that.