Let $F$ be any topological space. In many books, for example in http://pi.math.cornell.edu/~hatcher/VBKT/VBpage.html, it is said that a continuous map (characteristic map) (where $Homeo(F)$ has the compact-open topology) $$\phi \colon S^{n-1} \to Homeo(F)$$ define a fiber bundle $\xi_{\phi}$ over $S^n$ by gluing two trivial bundles with fiber $F$ over two hemispheres $D_+$ and $D_-$, whose total space is $$E = ((D_+ \times F) \coprod (D_- \times F)) / R$$ where the equivalence relation is for $x \in D_+ \cap D_- = S^{n-1}$ and $y \in F$ $$ (x,y) \in (D_+ \times F) \sim (x,\phi(x)(y)) \in D_- \times F).$$ In fact, Hatcher and others write that for vector bundles but here it doesn't matter.

I agree with them if you enlarge both hemispheres so that their intersection is an open equatorial zone. But if you use $D_+$ and $D_-$, I don't understand why $\xi_{\phi}$ is a fiber bundle.

The gluing theorem works for an open covering of the base, but here it is a closed covering, and their intersection iso to $S^{n-1}$ is also closed.

I don't see how the local triviality of this $\xi_{\phi}$ is obtained for $x \in S^{n-1}$. Strictly speaking, for me it is not a fiber bundle...


1 Answer 1


I doubt that is true for arbitrary spaces $F$, but it is true for locally compact $F$. This covers most relevant cases.

In the sequel let $F$ be locally compact. The exponential law tells us that continuous maps $\phi : S^{n-1} \to Homeo(F)$ can be identified with continuous maps $\phi' : S^{n-1} \times F \to F$ such that each $\phi'(x,-) : F \to F$ is a homeomorphism. Note that these maps can be identified with bundle isomorphims $\psi : S^{n-1} \times F \to S^{n-1} \times F$ : To $\phi'$ associate $\psi(x,y) = (x,\phi'(y))$ and to $\psi$ associate $\psi_F = p_F \circ \psi$ with projection $p_F : S^{n-1} \times F \to F$.

Also observe that in many cases one does not work with $Homeo(F)$, but with a subset $S \subset Homeo(F)$. For example, if $F = \mathbb{R}^m$, then one usually takes $S = GL(\mathbb{R}^m)$.

Gluing two trivial bundles over $D_\pm$ along $\psi$ is obvious and yields a space $E_\psi$ with projection $\pi :E_\psi \to S^n$. We want to show that $(E_\psi,\pi)$ is a fiber bundle. It remains to show local triviality around an arbitrary point $x \in S^{n-1}$. Let $U = S^n \setminus \{ northpole, southpole \}$. There is a canonical retraction $r : U \to S^{n-1}$. Define $\mu : U \times F \to \pi^{-1}(U)$ by $$\mu(x,y) = \begin{cases} [x,y] & x \in D_+ \cap U \\ [x,\psi_F(r(x),y)] & x \in D_- \cap U \end{cases} $$ This is a well-defined continuous map since on $S^{n-1} = D_+ \cap D_-$ we have $r(x) = x$ and $[x,y] = [x,\psi_F(x,y)]$. An inverse $\lambda : \pi^{-1}(U) \to U \times F$ for $\mu$ is given by $$\lambda([x,y]) = \begin{cases} (x,y) & x \in D_+ \cap U \\ (x,\psi^{-1}_F(r(x),y)) & x \in D_- \cap U \end{cases} $$ Note that it is induced by a continuous map $\lambda' : (D_+ \cap U) \times F \coprod (D_- \cap U) \times F \to U \times F$, i.e. is itself a continuous map.

  • $\begingroup$ OK, but why the restriction to F locally compact ? You need it here to get the exponential law a bijection, but why do you really need it here ? it seems you could do the same using directly $\phi$ instead of your $\Psi_F$ $\endgroup$
    – ychemama
    Feb 8, 2019 at 15:59
  • $\begingroup$ We have to know that $\phi' : S^{n-1} \times F \to F$ is continuous if $\phi$ is continuous. This requires $F$ locally compact. $\endgroup$
    – Paul Frost
    Feb 8, 2019 at 16:05
  • $\begingroup$ But in fact, if you ask that $\phi : S^{n-1} \to Homeo(F)$ be such that the map $S^{n-1} \times F \to F$, $(x,y) \mapsto \phi(x)(y)$ is continuous, your demo will still work. It is true that the supplementary condition is met if $F$ is locally compact", but it may still be satisfied in some larger context. $\endgroup$
    – ychemama
    Feb 8, 2019 at 16:22
  • $\begingroup$ I doubt that this is true for non-locally compact $F$. Perhaps it is worth to ask another question concerning this point? $\endgroup$
    – Paul Frost
    Feb 8, 2019 at 17:23
  • $\begingroup$ As I said: Perhaps it is worth to ask a question whether the condition of local compactness is necessary. I do not know the answer. $\endgroup$
    – Paul Frost
    Feb 8, 2019 at 23:23

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