Victor Dods
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 Aug28 awarded Yearling Aug6 awarded Scholar Aug6 awarded Supporter Aug6 accepted Image of commutative diagram is commutative under functor? Aug6 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. Aug6 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. Aug6 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? Aug6 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? Aug6 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 $y$). Aug6 asked Image of commutative diagram is commutative under functor? Jul20 awarded Revival Jul20 comment Product of Riemannian manifolds? Nice, that's a concise way to put it. Jul20 answered Product of Riemannian manifolds? Jul20 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? Jul19 awarded Student Jul19 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). Jul19 asked Alternate pullback bundle construction Jul19 awarded Teacher Jul19 answered The pullback $F^\ast :T^*N \rightarrow T^*M$ is a smooth bundle map