# What is an isomorphism of fiber bundles?

I am studying Algebraic Topology and struggling with isomorphism of fiber bundles, which is not explicitly defined by the professor. So what I am looking for is an explicit defintion of this.

The definition of fiber bundles he uses is as follows:

A continuoius map $$p:E\rightarrow B$$ between topological spaces is a fiber bundle if for every $$x$$ in $$B$$ there exist

i) an open neigbhourhood $$U$$ of $$x$$ in $$B$$

ii) A topological space $$F$$

iii) a homeomorphism $$h:p^{-1}(U)\rightarrow U\times F$$

such that $$p=\pi_U\circ h$$ where $$\pi_U$$ is the projection onto $$U$$.

Note that the prof is vehement about not using the term "fiber bundle" for a "total space".

Now, I know there is a category theory approach to this definition which some people have told me is simpler. But as someone who has little knowledge of category theory, I would appreciate a topological definition of isomorphism of fiber bundles and perhaps some correponding intuition.

• It's essentially just a homeomorphism of the total spaces which preserves the additional structure given by the bundle projection, meaning that if $E,F$ are the total spaces, then taking a point in $E$ and projecting it is the same as first applying the isomorphism and then projecting from $F$. Nov 6, 2021 at 12:00
• The category theoretical notion of isomorphism is indeed simple: it is a morphism that has an inverse. Nov 6, 2021 at 12:04
• Depending on the context, I can think of two notions of isomorphism. The first is the one @Vercassivelaunos mentions. But I one can also allows diffeomorphisms of the base. That is, in the alternative notion, an isomorphism is a pair $(f,g)$ where $f$ is a diffeo of the total space, $g$ is a diffeo of the base space, and $f$ and $g$ are required to fit into the obvious commutative diagram. The special case $g = Id$ gives the first notion. Nov 6, 2021 at 18:26

Two fiber bundles $$p_1:E_1\rightarrow B$$ and $$p_2:E_2\rightarrow B$$ with the same base space $$B$$ are isomorphic if ther exists a homeomorphism $$g:E_1\rightarrow E_2$$ such that $$p_2\circ g=p_1$$ Then, $$g$$ is called an isomorphism of fiber bundles $$p_1$$ and $$p_2$$.