How can I construct envelop of unity? Given a topological space $B$ and an open covering $\{ U_i \}_{i\in I}$ of $B$ with a partition of unity $\{ \varphi_i \}_{i \in I}$ such that $\operatorname{supp}(\varphi_i) \subset U_i$ $\forall i$, how can I construct a family of maps $\{ \eta_j \}_{j \in  J}$ with the properties that $\{\operatorname{supp}(\eta_j)\}$ is locally finite, $\forall j \exists i$ such that $\operatorname{supp}(\eta_j)\subset U_i$ and $\forall b \in B$ $\max_{j\in J}(\eta_j(b))=1$?
I found this problem studying the homotopical classification of fibre bundle in Husemoller's "Fibre Bundle".
 A: The maximum of a finite family of continuous functions is continuous. Since the supports of the $\varphi_i$ are a locally finite family, this implies that
$$\varepsilon := \sup_{i\in I} \varphi_i$$
is a continuous function on $B$, and since $\sum_i \varphi_i \equiv 1$, we have $\varepsilon(x) > 0$ for all $x \in B$.
Now define
$$\eta_i(x) := \min \left\lbrace 1, \frac{\varphi_i(x)}{\varepsilon(x)}\right\rbrace.$$
$\eta_i$ is continuous, and $\operatorname{supp} (\eta_i) = \operatorname{supp} (\varphi_i)$.
We have $\max\limits_{i \in I} \eta_i(x) = 1$ for all $x \in B$:
Since all $\eta_i$ are bounded above by $1$, it is clear that the maximum is at most $1$.
Since the family of supports is finite, $x$ has a neighbourhood $V$ that intersects only finitely many of the $\operatorname{supp} (\varphi_i)$, say for $i \in I_0$. Let $i_0 \in I_0$ be an index with $\varphi_{i_0}(x) = \max\limits_{i \in I_0} \varphi_i(x)$. Then $\varepsilon(x) = \varphi_{i_0}(x)$, and hence $\eta_{i_0}(x) = 1$, so the maximum is at least $1$ at every point.
