Short exact sequence of $K$ functor; applying Tietze extension I have questions in  page 52 of Hatcher's, for the proof of Proposition 2.9

If $X$ is a compact Hausdorff space and $A \subseteq X$ is a closed subspace, then the inclusion and quotient maps $$A \xrightarrow{i}X \xrightarrow{q} X/A$$ induce homomoprhisms $$\tilde{ K}(X/A) \xrightarrow{q^*} \tilde{ K}(X) \xrightarrow{i^*} \tilde{ K}(A)$$

He makes the following claim. 
In showing that $\ker i^* \subseteq im \, q^*$, he applied the Tietze extension theorem. 

$E$ is trivial over $A$ implies there are sections $s_i$. Choose a cover of $A$ by open sets $U_j$ in $X$ over each of which $E$ is trivial over. We may regard restrictions $s_{ij}:A \cap U_j \rightarrow E$ as maps into a single fiber. i.e. 
  $$ s_{ij} :A \cap U_j \rightarrow \Bbb C^n$$ 
  By the Tietze extension theorem we may extend this to $s_{ij} :U_j \rightarrow \Bbb C^n$. 

But why can we apply? Tietze extension theorem requires $U_j$ to be normal, it is only a open  subset of a compact Hausdorff space. 
 A: You are right, Hatcher is again a little imprecise, and in the present form it does not work. Let us repair it.
For each $x \in A$ choose an open neighborhood $V(x)$ in $X$ over which $E$ is trivial. Next choose a neigborhood $U(x)$ such that $\overline{U(x)} \subset V(x)$. Then the $U(x)$ cover $A$. Now Tietze's extension theorem applies to give extensions $s'_{ix} : \overline{U(x)} \to \mathbb{C}^n$ of $s_i \mid_{A \cap \overline{U(x)}} : A \cap \overline{U(x)}  \to \mathbb{C}^n$. Then $s_{ix} = s'_{ix} \mid_{U(x)} : U(x) \to \mathbb{C}^n$ extends $s_i \mid_{A \cap U(x)} : A \cap U(x)  \to \mathbb{C}^n$. Now you are in the adequate position to proceed like Hatcher.
Up to this point the argument works for any normal $X$. In the following step a partition of unity is used which requires $X$ paracompact.
A: Just note the following facts:

*

*A compact Hausdorff space is normal, thus completely regular;

*Any subspace of completely regular space is completely regular;

*On completely regular space, the Urysohn lemma applies, and hence the Tietze extension theorem.

