# B-Isomorphic? (Husemoller, Fiber Bundles). Illustrated diagrams included.

Wikipedia gives the definition of a bundle map as the arrow $$\left( \xrightarrow{\quad \varphi \quad } \right)$$ in the commuting diagram below:

Hussemoller gives the definition of a bundle map as the arrow (note the ordered pair) $$\left( \xrightarrow{\quad \left(u, f \right) \quad } \right)$$ in the diagram below:

(i)

Are these definitions distinct, or are they just different preferences for stating the same concept?

(ii)

Is $$(f, u)$$ a functor? If it's a functor, doesn't there need to be a map $$\big( p \xrightarrow{\quad} p' \big)$$? Why is it sufficient to only represent the map between bundles by mapping the objects in the category?

Now, the wikipedia article doesn't define isomorphisms, and Hussemoler doesn't give any more diagrams. From Husemoller (bundle isomorphism): "From general properties in a category, a bundle morphism $$(E, p, B) \xrightarrow{\quad (u, f)\quad} (E', p', B')$$ is an isomorphism $$\iff$$ there exists a morphism $$(E, p, B) \xleftarrow{\quad (u', f')\quad} (E', p', B')$$ with $$f f' = 1_B$$, $$f'f = 1_{B'}$$, $$u'u = 1_E$$, and $$uu' = 1_{E'}$$."

(iii)

Does this diagram I drew (which commutes) show his bundle isomorphism?

Now, for most unclear part of Husemoller's text:

"A space $$F$$ is the fibre bundle $$(E, p , B)$$ provided every fibre $$p^{-1}$$ for $$b \in B)$$ is homeomorphic to $$F$$. A bundle $$(E, p, B)$$ is trivial with fibre $$F$$ provided $$(E, p, B)$$ is B-isomorophic to the product bundle $$(B \times F, p , B)$$."

This is the first time in his entire text that Husemoller uses the terms trivial and B-isomorphic. The Wikipedia article and Husemoller both give similar diagrams for a B-homomorphism (here is Wikipedia's for an M-homomorphism):

so I'm going to take a guess and ask...

(iv)

Would this (commuting) diagram I drew represent what he means by both bundles being B-isomorphic and describe the situation that he's laying out?

As always, thanks in advance for the help.

The definitions are equivalent: You can always recover $$f$$ from $$\varphi$$ by noting that the definition implies that $$\varphi$$ takes fibres to fibres, so if $$\varphi(E_x) \subseteq F_y$$ where $$E_{-}, F_{-}$$ are the fibres over the respective points, then we must have $$f(x) = y$$.
Who says that $$(u, f)$$ is a functor? It is a morphism in the category of bundles. This explains Husemollers remark about isomorphisms. The diagram you drew is per se of the right shape, but to capture the definition you need to specify that the top and bottom composites be the identity in both directions.
For the last part, you've almost got it right: The projection map to $$B$$ on the right should be $$\pi_X\colon X \times F \to X$$, the projection to the first factor. Alternatively, you could express the same fact by saying that there is a bundle isomorphism $$(E, p, B) \overset{(u, \mathrm{id}_B)}{\to} (B \times F, \pi_X, B)$$ (more generally, a $$B$$-homomorphism is just a bundle map covering $$\mathrm{id}_B$$).