Consider some kind of bundles, for instance vector bundles or fibre bundles with a certain structure group, such that there are bundle morphisms and pull-backs.

Let then $F(X)$ denote the isomorphism classes of bundles over $X$, this is a contravariant functor by virtue of the pull-back construction. Assume further that $F(.)$ satisfies the conditions of Brown's representabiliy theorem, which is usually the case in such a setting.

So we find a space $B$ such that there is a natural isomorphism $F(X) \cong [X,B]$. Let $ \gamma$ denote the bundle, which corresponds to [id_B] under this isomorphism.

Is $\gamma$ then a universal bundle?

I can prove that for every bundle there is a morphism to $\gamma$ and that the base map of this morphism is unique up to homotopy. It is however not clear to me, why the morphism itself is unique up to homotopy.

If we consider for instance orientable or oriented vector bundles, then $\gamma$ has a self-isomorphism induced by sending every vector to its negative and in odd dimension this is not orientation preserving, so the automorphism can not be homotopic to the identity as a morphism. Therefore $\gamma$ cannot be a universal bundle.

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    $\begingroup$ I think you are misunderstandiing the definition of universal bundle. Usually only what you call the "base morphism" is required to be unique: as you correctly note the universal vector bundle of rank $n$ does have automorphisms. Basically the definition of universal bundle is rigged so to correspnd to a natural isomorphism $F(X)\cong [X,B]$. $\endgroup$ Aug 27, 2016 at 11:27
  • $\begingroup$ @DenisNardin Ok, then this is the reason for my confusion. I took the definition from arxiv.org/abs/math/0105047 Where Rudyak defines a universal fibre bundle like this on page 22. Theorem-Definition 2.2 $\endgroup$ Aug 27, 2016 at 11:37
  • $\begingroup$ Sorry, I unintentionally posted it before I was finished. What he requires then has to be a much stronger condition. I am not even sure, whether it is still true like this $\endgroup$ Aug 27, 2016 at 11:43
  • $\begingroup$ I think Proposition 2.3 states that the morphism is unique up to homotopy. $\endgroup$ Aug 27, 2016 at 11:47
  • $\begingroup$ Right, that sounds stronger. In general you'll have to twist the inclusion $\xi_A\to \xi$ by an automorphism of $\xi_A$. See for example remark 2.4: he seems to say that all trivializations of a trivial bundle are homotopic, but that's simply false: the set of trivialization up to homotopy is isomorphic to the set $[X,Top_N]$ $\endgroup$ Aug 27, 2016 at 11:51

2 Answers 2


There seems to be a little confusion here. Let me try to be elementary. For any type of bundle (or fibration) closed under pullbacks, a bundle $\gamma \colon E\to B$ of the specified type is defined to be universal if pullback of $\gamma$ along maps $f\colon X \to B$ induces a natural isomorphism $[X,B]\to F(X)$, where $F(X)$ is the set of isomorphism classes of bundles of the speciified type, and naturality refers to maps that preserve whatever structure that class of bundles has. That is the (or a) standard definition in algebraic topology; here one might want to restrict the $X$ to have the homotopy types of CW complexes if one has classical insecurities. Then, by definition, $\gamma$ is a universal bundle and is the bundle which corresponds to $[id_B]$ under the isomorphism. When one constructs $\gamma$ by Brown representability, one is representing the functor $F$, so getting the required bundle $\gamma$. The fallacy in the last paragraph of the question is that one cannot take orientable bundles but one can take oriented bundles, so that a map of bundles must preserve the orientation. Then the supposed contradiction disappears.

I don't want to go into details on the definitions in https://arxiv.org/abs/math/0105047, but they are fairly non-standard from the usual point of view of algebraic topology. The idea of 2.2 seems to be that there should be no obstructions to extending bundles, as should follow or be equivalent to the contractibility of the total space of $\gamma_{TOP}^n$. There is no uniqueness claim in the extendability condition, even up to bundle homotopy. Then 2.3 gives the uniqueness of the classifying map (not classifying morphism, which is a morphism of bundles, not of base spaces). In 2.4 he is saying explicitly that classifying maps (base space level) are unique up to homotopy but that classifying morphisms (bundle level) need not be unique up to homotopy, as was illustrated (or meant to be illustrated) by the example of trivial bundles. Judging by the number of typos, that paper seems to have been written in haste, but it doesn't seem obviously wrong.

Incidentally, although the ordinary cohomology of $BTop$, the colimit of the $BTop(n)$, is well understood, the mod $p$ cohomologies of the $Top(n)$ are still not fully computed as far as I know.

  • $\begingroup$ Thank you for clarifying which definition is usually used. As you mentioned it is meant to make the classifying map unique up to homotopy. The definition given in the arXiv article implies that the classifying morphism is unique up to homotopy (define a morphism on $X\times \{0,1\}$ and extend it to $X\times [0,1]$). It turns out that the definitions are often equivalent, see my answer below. Btw I read your book on simplicial objects in algebraic topology for my bachelor thesis learned a lot, thank you for writing $\endgroup$ Nov 6, 2016 at 21:22

By Assertation 8.3 in Essay IV of "Foundational Essays on Topological Manifolds, Smoothings, and Triangulations" by Kirby and Siebenmann, the following two definitions of a universal bundle $\gamma$ (over a space $BG$) are equivalent "for most types of bundles":

(*) For every bundle $\xi$ over a base space $B$, every closed subset $C \subset B$ and every morphism $\varphi:\xi_{|U} \to \gamma$, where $U$ is a neighbourhood of $C$, there is a morphism $\psi: \xi \to \gamma$ such that $\psi_{|C} = \varphi_{|C}$.

(**) For every bundle $\xi$ there is a classifying map $f:B \to BG$ such that $\xi \cong f^*\gamma$ and this $f$ is unique up to homotopy.

The first property implies that the classifying morphism $\xi \to \gamma$ is unique up to homotopy (take $B = X \times [0,1]$, $C=X \times \{0,1\}$ and $U=X \times [0,1/3) \cup (2/3,1]$), whereas the second property only requires that the base map of such a morphism is unique up to homotopy.

Often only (**) is proven (for example for vector bundles in Milnor-Stasheff "Characteristic Classes"), but by the statement all these (**)-universal bundles are as well (*)-universal. This implies for instance that every automorphism of the universal bundle $\gamma \to \gamma$ is homotopic to the identity.

As it was mentioned in one of the comments above, I also want to regard the special case of the trivial bundle $\xi$ over a base $X$. For simplicity, assume $X$ is the point $*$. If we fix a base map $* \to BG$, then there are $[*,G]=\pi_1 G$ many morphisms over it, up to homotopy with this fixed base map. (Here $G$ is the structure group of the type of bundles, which we regard. So $Gl_n$ for vector bundles.)

Note however that $\pi_0 G = \pi_1 BG$, so every element $g$ of $[*,G]$ corresponds to a loop $c:[0,1] \to BG$. If we now regard a morphism $\psi: \xi \times [0,1] \to \gamma$, which has $c$ as its base map, then the morphisms $\psi_{|\xi \times \{0\}}$ and $\psi_{|\xi \times \{1\}}$ should differ by $g$, so $$ \psi_{|\xi \times \{1\}} = \psi_{|\xi \times \{0\}} \cdot g.$$ This means that if we do not fix the base map of the classifying morphism $\xi \to \gamma$, there is really only one up to homotopy.

I interpret "for most types" as: We can take every kind of bundle (fibre bundle, microbundle etc.) but we should not regard bundles with an orientation or a framing. Or, at least, then we have to require that the morphisms preserve this extra structure. Otherwise, the oriented line bundles would yield a counter-example.


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