Confusion about vector bundle extension lemma from Atiyah's K-Theory

Question

Lemma 1.4.2 from Atiyah's K-theory book (p16-17): Let $Y$ be a closed subspace of a compact Hausdorff space $X$, and let $E,F$ be two vector bundles over $X$. If $f:E\vert_Y \to F \vert_Y$ is an isomorphism, then there exists an open set $U$ containing $Y$ and an extension $f : > E\vert_U \to F \vert_U$ which is an isomorphism.

Proof: $f$ is a section of $Hom(E\vert_y,F\vert_y)$ and thus extends to a section of $Hom(E,F)$. Let $U$ be the set of those points for which this map is an isomorphism. Then $U$ is open and contains $Y$.

I understand we can extend $Hom(E\vert_y,F\vert_y)$ to $Hom(E,F)$ since $Hom(E,F)$ is a vector bundle and we can apply the bundle form of the Tietze extension theorem. However, I don't understand how the set $U$ defined in the proof is open. By hypothesis we only have that $f$ is an isomorphism over $U$ which is a closed subspace, so it seems like the set of points making $Hom(E,F)$ an isomorphism would be equal to $Y$ which is closed. Any clarification on how $U$ is open would be helpful.

Answer 1

Not a serious proof.

The vector bundle isomorphism $f:E\vert_Y\to F\vert_Y$ provides a map (not so sure how) $$t:Y\to GL_n(F)\subset Mat_n(F)=F^{n^2}$$ By Tietze extention theorem we can extend $t$ to $$\tilde t: X\to Mat_n(F).$$ Let $K=\tilde t^{-1}(\Delta)$ where $\Delta=Mat_n(F)-GL_n(F).$ Then $K$ is closed, $U=X-K$ is open and $Y\subset U.$ Furthermore $\tilde t\vert_U$ gives an extension of the vector bundle isomorphism $f$ on $U$.

Answer 2

The section $g \in \Gamma(Hom(E, F))$ that we obtain by extending $f$ is a bundle map $E \to F$. Now we can define the following function \begin{align} h : & X \to \mathbb{C} \\ & x \mapsto \det(g_x : E_x \to F_x) \end{align} This is continuous (being continuous is a local property so we can assume both $E$ and $F$ are trivial and for trivial bundles it is clearly continuous) and thus $U = h^{-1}(\mathbb{C} - \{0\})$ is open.