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seen Aug 6 '12 at 7:48

Aug
28
awarded  Yearling
Aug
6
awarded  Scholar
Aug
6
awarded  Supporter
Aug
6
accepted Image of commutative diagram is commutative under functor?
Aug
6
comment Image of commutative diagram is commutative under functor?
Ah, ok. In my example, there were no nontrivial commutativity conditions in the domain category, and therefore none were required in the image. Thanks.
Aug
6
comment Image of commutative diagram is commutative under functor?
Yes, that's right. Why though are there two copies of each of $a,b,c,d$ in the second picture? I assume the intent is to imply that certain arrows aren't to be composed when talking about the commutativity of the diagram. But if this is the case, then the notion of commutativity in this case seems too weak to be useful.
Aug
6
comment Image of commutative diagram is commutative under functor?
The square part consisting of the arrows $\phi, \psi, \theta, \omega$ is a diagram in $D$ which is the image of $C$ under $F$. Or does this fail to be a diagram at all, because it is not closed under composition of arrows?
Aug
6
comment Image of commutative diagram is commutative under functor?
I think that's not quite right -- the endpoints of the arrows for $f,g,x,y,\phi,\psi,\omega,\theta$ are backwards. I think I see what you're getting at by showing two copies of each of $a,b,c,d$ is that the image of a diagram under a functor is distinct from a diagram in the codomain of the functor. If this is the case, then what is the distinction exactly? Using some sort of pullback category?
Aug
6
comment Image of commutative diagram is commutative under functor?
While it's certainly true that if you have a commuting square diagram in the domain category then its image is a commuting square diagram in the codomain category, what happens when there is no such square (or even triangle) in the domain category? In my example, the composition $F(y) \circ F(f)$ is well-defined, but not the composition $y \circ f$ (since the codomain of $f$ is not equal to the domain of $y$).
Aug
6
asked Image of commutative diagram is commutative under functor?
Jul
20
awarded  Revival
Jul
20
comment Product of Riemannian manifolds?
Nice, that's a concise way to put it.
Jul
20
answered Product of Riemannian manifolds?
Jul
20
comment Alternate pullback bundle construction
Identifying the naturally isomorphic pullback bundles in the category of bundles seems like the way to go, but I can't think of how to do this in the category of bundles -- it seems like one must talk only about the subcategory of pullback bundles (assuming this is actually a subcategory); put an equivalence relation on this category which relates isomorphic bundles in the sense described above. OR, is there a more natural way to do this in the full category of bundles?
Jul
19
awarded  Student
Jul
19
comment Alternate pullback bundle construction
My thought is that the submanifold-of-direct-product construction could still be used as "hidden implementation details", over which an equivalence relation is imposed to identify the isomorphic spaces, and then somehow translate into this scheme the natural morphisms/maps one gets from the fact that the pullback bundle is defined as a submanifold of a direct product (e.g. restrictions of projections onto each factor, etc).
Jul
19
asked Alternate pullback bundle construction
Jul
19
awarded  Teacher
Jul
19
answered The pullback $F^\ast :T^*N \rightarrow T^*M$ is a smooth bundle map