Clutching function for $S^n$: why a vector bundle?

Question

On Page 22 of Hatcher's $K$ Theory we make the following construction

For $n,k \in \mathbb N \setminus \{ 0\}$. Let $f:S^{k-1} \rightarrow GL_n(\mathbb R)$ be a continuous map. Define $$E_f:= (D_{-}^k \times \mathbb R^n \sqcup D_+^k \times \mathbb R^n ) / \sim $$ where $\sim$ is equivalence relation generated by $(x,v) \sim (x, f(x)v)$ for $x \in D^k_{-} \cap D^k_+$.

From the map induced from projection $\pi:E_f \rightarrow S^k$, Hatcher claims this is a vector bundle. I don't understand his argument for the local trivialzation condition.

He says this is seen by

taking an equivalent in definition... a product $S^{k-1} \times (-\varepsilon, \varepsilon)$ with the map $f$...

May someone elaborate this in detail?

Answer 1

I initially misunderstood the notation $D^k_+$ and $D^k_-$. The problem is that these are closed sets, and the definition of local triviality involves open sets. Hatcher prefers to use this definition so the 'clutching function' is only defined on the equator, not a larger (but homotopy equivalent!) set.

Instead, take $D^k_+$ to mean an open set in the $n$-sphere including the north hemisphere and a band including the equator. In fact, you might as well just say "Let $D^k_+$ mean everything except the south pole." So let's do that. Similarly for $D^k_-$: define it as the sphere minus the north pole. Then their intersection is $S^k \setminus \{N, S\} \cong S^k \times \Bbb R \simeq S^{k-1}$. We choose a function $f: S^k \setminus \{N, S\} \to GL_n(\Bbb R)$ to serve as the transition function.

Now the formula for $E_f$ is identical to the formula you gave, and (tautologically) $E_f$ has canonical trivializations over $S^k \setminus \{N\}$ and $S^k \setminus \{S\}$.

---

If the question is why this is equivalent to the previous definition: Choose $f': S^{k-1} \to GL_n(\Bbb R)$ to be the restriction of $f$ above to the equator. There is a map $E_{f'} \to E_{f}$ given by sending (north hemisphere) x $\Bbb R^n$ to the corresponding subset in $D^n_+ \times \Bbb R^n$, and similarly for the southern hemisphere; these agree on the overlap (the equator) precisely because $f' = f|_{S^{k-1}}$.

Answer 2

I will show how we can pull a trivilization from an extension to the original definition. Given $f:S^{k-1} \rightarrow GL_n(\mathbb R)$, let $r: S^n \setminus \{N,S\} \rightarrow S^{n-1}$ be a retraction.

Then this induces a map $f':S^n \setminus \{N,S\} \rightarrow GL_n(\mathbb R)$ such that $$f'(x)=f(r(x)), \quad \quad (*)$$ Let us denote $\hat{E}_f$ as the bundle induced from $f'$, which Mike Miller has proven above.

---

There are trivial inclusions, $$D^n_- \times \mathbb R^n \rightarrow S^n \setminus \{ N\} \times \mathbb R^n \rightarrow \hat{E}_f, $$
$$D^n_+ \times \mathbb R^n \rightarrow S^n \setminus \{S\} \times \mathbb R^n \rightarrow \hat{E}_f $$ and under quotient relation, inducing $$ i: E_f \rightarrow \hat{E}_f$$

Note that there are retractions $S^n \setminus \{ S \} \rightarrow D^n_+$. Thus, this induces map $$ S^n \setminus \{ S\} \times \mathbb R^n \rightarrow D^n_+ \times \mathbb R^n \rightarrow E_f $$

A similar map for the sphere with north pole taken induces $$ i: S^n \setminus \{S \} \times \mathbb R^n \sqcup S^n \setminus \{N \} \times \mathbb R^n \rightarrow E_f. $$ now note that the points $(x,v), (x,f'(x)^{-1}v)$ gets identified for $x \notin S^{n-1}$, by $(*)$. $$ (x,v) \mapsto (r(x),v) \mapsto [r(x),v]$$ $$ (x,f'(x)^{-1}v) \mapsto (r(x),f'(x)^{-1}v) \mapsto [r(x), f(r(x))f'(x)^{-1}v] = [r(x),v]$$ Thus, this induces a continuous map $$ i^{-1}:E_{f'} \rightarrow E_f $$ which by element chase, can be seen to be a clear inverse of $E_f$. As the map is linear on each fiber, we can utilize the localization in $\hat{E}_f$ for that of $E_f$.