# Fiber Bundles over spheres

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...

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.

• 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 $\Psi_F$ Feb 8, 2019 at 15:59
• We have to know that $\phi' : S^{n-1} \times F \to F$ is continuous if $\phi$ is continuous. This requires $F$ locally compact. Feb 8, 2019 at 16:05
• 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. Feb 8, 2019 at 16:22
• I doubt that this is true for non-locally compact $F$. Perhaps it is worth to ask another question concerning this point? Feb 8, 2019 at 17:23
• 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. Feb 8, 2019 at 23:23