Showing a map $f:E \rightarrow \Bbb R^\infty$ is continuous. 
Suppose for $i \in \Bbb N$, we have  continuous maps $g_i:E \rightarrow \Bbb R^n$ satisfying local finiteness proeprty i.e. for each $x\in E$ exists nhood $U_x$ such that only finitely many $g_i$ are nonzero. 
Define $f:E \rightarrow \Bbb R^\infty$, $\Bbb R^\infty$ being direct limit of inclusions $\Bbb R^i$, as the composition, 
$$ E \xrightarrow{f'} \bigsqcup \Bbb R^n \xrightarrow{q} \Bbb R^\infty$$ 
  where $f':x \mapsto (H_i(x))_{i=1}^\infty$, and $\bigsqcup \Bbb R^n$ is coproduct in category of topological space, and $q$ is the quotient map. 

I have two questions, (i) is the finiteness property necessary, and (ii) is it true that this map is continuous? 

My argument:
It suffices to show $f'$ is continuous. 
Let $U \subseteq \bigsqcup \Bbb R^n$ be open nhood of $f'(x)$ and $U_x$ a local finite nhood of $x$ in $E$.
So it suffices to work locally, Let $U_i:=  U \cap \Bbb R^i$. Then, $$f'^{-1}(U) \cap U_x= \bigcap_{i=1}^\infty (g_i^{-1}(U_i) \cap U_x) = \bigcap_{i=1}^{k} g_{n_i}^{-1}(U_{n_i}) \cap U_x$$
is non empty open . 

This question is from Theorem 2.10, pg 15, line 7-10 of proof. 
 A: The author starts with a countable locally finite open cover $\{ U_i \}$ and constructs continuous maps $H_i : E \to \mathbb{R}^n$ such that $H_i(x) = 0$ for $x \notin p^{-1}(U_i)$. Then he defines
$$f : E \to \oplus_{i=1}^\infty \mathbb{R}^n = \mathbb{R}^\infty, f(x) = (H_i(x)) .$$
$\oplus_{i=1}^\infty \mathbb{R}^n$ is not the coproduct in category of topological spaces, it is the direct sum of vector spaces. That is, $\oplus_{i=1}^\infty \mathbb{R}^n = \{ (y_i) \in \prod_{i=1}^\infty \mathbb{R}^n \mid y_i \ne 0 \text{ only for finitely many } i \}$. It receives the subspace topology from the infinite product $ \prod_{i=1}^\infty \mathbb{R}^n$ (which has of course the product topology).
Note that we may identify $\oplus_{i=1}^\infty \mathbb{R}^n$ with $\oplus_{i=1}^\infty \mathbb{R} = \{ (t_i) \in \prod_{i=1}^\infty \mathbb{R} \mid t_i \ne 0 \text{ only for finitely many } i \}$. The latter is the standard definition of $\mathbb{R}^\infty$.
Regarding $f$ as a map into $\prod_{i=1}^\infty \mathbb{R}^n$, continuity is obvious. It remains to show that $f(E) \subset \mathbb{R}^\infty$. This means that for each $x$ only finitely many $H_i(x) \ne 0$. But this is clear because $\{ U_i \}$ is locally finite. Given $x \in E$, there are only finitely many $i$ such that $p(x) \in U_i$. For all other $i$ we have $p(x) \notin U_i$, i.e. $x \notin p^{-1}(U_i)$ and hence $H_i(x) = 0$.
