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Will Sawin
Will Sawin
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I think you're confused. The issue in )(1) is not about strict functors. Instead, it is about what it means for a trio of functors to be commutative. If $f: A \to B$, $g: B \to C$, and $h: A \to C$ form a commutative triangle, that means $g \circ f = h$ - they are the same functor. But even if these are strictly equivalent functors, there are many different choices for the isomorphism between $g \circ f$ and $h$. Indeed, there are many different choices for the isomorphism between a functor and itself. In that case you can choose the identity, but that won't have meaning here.

A commutative triangle of functors should be a seen as trio $f,g,h$, with an isomorphism $g \circ f = h$. This is what gets you a morphism of groups, and this is the appropriate notion of a morphism of Tannakian categories.

In fact, the different functors from $Rep_G$ to itself are all the same strict functor, because they act the same on the set of representations (assuming we take the set of representations to have one element for each isomorphism class.)

I think you're confused. The issue in )1) is not about strict functors. Instead, it is about what it means for a trio of functors to be commutative. If $f: A \to B$, $g: B \to C$, and $h: A \to C$ form a commutative triangle, that means $g \circ f = h$ - they are the same functor. But even if these are strictly equivalent functors, there are many different choices for the isomorphism between $g \circ f$ and $h$. Indeed, there are many different choices for the isomorphism between a functor and itself. In that case you can choose the identity, but that won't have meaning here.

A commutative triangle of functors should be a seen as trio $f,g,h$, with an isomorphism $g \circ f = h$. This is what gets you a morphism of groups, and this is the appropriate notion of a morphism of Tannakian categories.

In fact, the different functors from $Rep_G$ to itself are all the same strict functor, because they act the same on the set of representations (assuming we take the set of representations to have one element for each isomorphism class.)

I think you're confused. The issue in (1) is not about strict functors. Instead, it is about what it means for a trio of functors to be commutative. If $f: A \to B$, $g: B \to C$, and $h: A \to C$ form a commutative triangle, that means $g \circ f = h$ - they are the same functor. But even if these are strictly equivalent functors, there are many different choices for the isomorphism between $g \circ f$ and $h$. Indeed, there are many different choices for the isomorphism between a functor and itself. In that case you can choose the identity, but that won't have meaning here.

A commutative triangle of functors should be a seen as trio $f,g,h$, with an isomorphism $g \circ f = h$. This is what gets you a morphism of groups, and this is the appropriate notion of a morphism of Tannakian categories.

In fact, the different functors from $Rep_G$ to itself are all the same strict functor, because they act the same on the set of representations (assuming we take the set of representations to have one element for each isomorphism class.)

Source Link
Will Sawin
Will Sawin
  • 148.4k
  • 9
  • 324
  • 563

I think you're confused. The issue in )1) is not about strict functors. Instead, it is about what it means for a trio of functors to be commutative. If $f: A \to B$, $g: B \to C$, and $h: A \to C$ form a commutative triangle, that means $g \circ f = h$ - they are the same functor. But even if these are strictly equivalent functors, there are many different choices for the isomorphism between $g \circ f$ and $h$. Indeed, there are many different choices for the isomorphism between a functor and itself. In that case you can choose the identity, but that won't have meaning here.

A commutative triangle of functors should be a seen as trio $f,g,h$, with an isomorphism $g \circ f = h$. This is what gets you a morphism of groups, and this is the appropriate notion of a morphism of Tannakian categories.

In fact, the different functors from $Rep_G$ to itself are all the same strict functor, because they act the same on the set of representations (assuming we take the set of representations to have one element for each isomorphism class.)