pullback square造句

例句与造句

1. Now this is where I'm confused; what is " the pullback of a pullback square "?
2. I guess the'pullback of a pullback square'is not meant to be assumed to be a pullback of the 2 commuting maps, otherwise the question would be done.
3. To me, a pullback is the limit of a diagram consisting of 2 morphisms with the same codomain ( as defined on the pullback, these would be gp _ 2, \, fp _ 1 ) : but then if we are simply taking a'pullback of a pullback square'to be the pullback of these 2 morphisms, then how could it possibly not be a pullback?
4. By considering the pullback square of f and T, which pulls back to an object U and morphism u : U \ to \ Omega and the unique morphism U \ to 1 say, and the pullback square of morphisms T and u to some object V ( and the composite of these 2 pullback squares,'joined'by u ), show that f \ circ f is the identity morphism : 1 _ { \ Omega }.
5. By considering the pullback square of f and T, which pulls back to an object U and morphism u : U \ to \ Omega and the unique morphism U \ to 1 say, and the pullback square of morphisms T and u to some object V ( and the composite of these 2 pullback squares,'joined'by u ), show that f \ circ f is the identity morphism : 1 _ { \ Omega }.
6. It's difficult to find pullback square in a sentence. 用pullback square造句挺难的
7. By considering the pullback square of f and T, which pulls back to an object U and morphism u : U \ to \ Omega and the unique morphism U \ to 1 say, and the pullback square of morphisms T and u to some object V ( and the composite of these 2 pullback squares,'joined'by u ), show that f \ circ f is the identity morphism : 1 _ { \ Omega }.
8. The first 2 parts of the problem ( which i've done ) are about a concatenation of 2 pullback squares ( http : / / www . dpmms . cam . ac . uk / ~ jg352 / pdf / CTSheet2-2011 . pdf Q2 ) : the third part of the problem asks you to deduce, from the fact that combining 2 of the pullbacks in the appropriate way ( e . g . concatenating the 2 small squares ) gives you another pullback, that " the pullback of a pullback square is a pullback ".
9. The first 2 parts of the problem ( which i've done ) are about a concatenation of 2 pullback squares ( http : / / www . dpmms . cam . ac . uk / ~ jg352 / pdf / CTSheet2-2011 . pdf Q2 ) : the third part of the problem asks you to deduce, from the fact that combining 2 of the pullbacks in the appropriate way ( e . g . concatenating the 2 small squares ) gives you another pullback, that " the pullback of a pullback square is a pullback ".
10. Really, so far the only ideas I have are to use the fact that the'composite'of 2 pullbacks in the way I've described is "'also "'a pullback square, and to use the fact that the'subobject classifier property'applies to all monomorphisms, therefore we could apply the property to f to find some unique corresponding morphism \ chi _ f : \ Omega \ to \ Omega to get a third pullback square and then perhaps use that somehow, but otherwise I've really been unable to make progress.
11. Really, so far the only ideas I have are to use the fact that the'composite'of 2 pullbacks in the way I've described is "'also "'a pullback square, and to use the fact that the'subobject classifier property'applies to all monomorphisms, therefore we could apply the property to f to find some unique corresponding morphism \ chi _ f : \ Omega \ to \ Omega to get a third pullback square and then perhaps use that somehow, but otherwise I've really been unable to make progress.