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.

  • 2
    $\begingroup$ 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 $F$. $\endgroup$ Nov 6, 2021 at 12:00
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    $\begingroup$ The category theoretical notion of isomorphism is indeed simple: it is a morphism that has an inverse. $\endgroup$
    – Arthur
    Nov 6, 2021 at 12:04
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    $\begingroup$ 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. $\endgroup$ Nov 6, 2021 at 18:26

1 Answer 1


I found the definition.

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 $p_2$.

(This is basically the definition given by Vercassivelaunos in the comments written explicitly.)


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