[The nLab page on partitions of unity](http://ncatlab.org/nlab/show/partition+of+unity) mentions the application of partitions of unity as a way to construct continuous maps to geometric realizations of simplicial spaces. However I often feel uncomfortable with the continuity of maps constructed in the realm of this example.


Let me discuss my uncertainty with the help of an explicit example.


Let $X$ be a space with an open cover $X=\bigcup\_{i\in I} U\_i$ and a partition of unity $\{f\_i\colon X\rightarrow\mathbb{R}\}\_{i\in I}$ subordinate to this cover and $\{a\_i\}\_{i\in I}$ a sequence of arbitrary real numbers. 


Consider the poset $(\mathbb{R},\le)$ as a category internal to Spaces with the usual Euclidian topology. The nerve of this is a simplicial space $N\_\bullet(\mathbb{R},\le)$. It has a geometric realization $||N\_\bullet(\mathbb{R},\le)||=B(\mathbb{R},\le)$, whereby I mean the so called "fat" geometric realization, i.e. I only factor out the face maps and not the degeneracies.


Now define a map $g\colon X\rightarrow||N\_\bullet(\mathbb{R},\le)||$ by mapping $x$ to the residue class of $((a\_{i\_0}\le...\le a\_{i\_k}),(f\_{i\_0}(x),...,f\_{i\_k}))\in N\_k(\mathbb{R},\le)$ where ${i\_0,...,i\_k}$ are exactly those indices where $f\_i(x)\neq0$.


This gives a well-defined map. Is this map always continuous? The choice of the non-zero $f\_i(x)$'s feels somehow uncomfortable to me. The values of $g$ seem to "jump" somehow.


However, my impression that many of the constructions summarized by the mentioned example of the nlab go along this example, so probably my feeling is wrong and I would be happy seeing somehow thin out my fog.




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Edit: The comment of Omar Antolín-Camarena and the answer of Oscar Randal-Williams have shown, that the example is not suitable to address my concerns. Let my give it another try by means of a modified example using the notation of the example before.


Let $Y\subseteq \mathbb{R}\times X$ a subspace. View $Y$ as a topological preordered set by using the standard ordering in the first coordinate, i.e. $(a,x)\le(b,y):\iff a\le b\text{ and } x=y$. Assume there are $a\_i\in \mathbb{R}$, such that $\{a\_i\}\times U\_i\subseteq Y$. Now define the map $$g\colon X\rightarrow ||N\_\bullet(Y)||$$ analogously by $$g(x)=[((a\_{i\_0},x)\le...\le(a\_{i\_k},x)),(f\_{i\_0}(x),...,f\_{i\_k}(x)],$$ where the indices $i\_0,...i\_k$ are exactly those, where $f\_i(x)\neq 0$. Now one does not have the freedom to take a bigger set of indices, because if $f\_i(x)=0$ it could be possible that $(a\_i,x)\notin Y$, so the suggestion of Oscar Randal-Williams and Omar Antolín-Camarena do not seem to work.



Continuous maps to fat geometric realizations of simplicial spaces
My first remark concerns Segal's (4.1). The technical details are key. Segal begins by choosing a *locally finite partition of unity* $f\_i$ subordinate to the cover $U\_i$. The definition of this term on the nLab seems to be non-standard, and probably does not make the statement true. 


A partition of unity $\{f\_i\}\_{i \in I}$ is usually called "locally finite" if the open cover $f\_i^{-1}((0,1])$ is locally finite, i.e. each point has an open neighbourhood intersecting only finitely many $f\_i^{-1}((0,1])$. 


Suppose then, in the notation of the question, that the partition of unity $\{f\_i\}\_{i \in I}$ is locally finite in the sense just described. Let $x \in X$ and $V\_x \ni x$ be an open neighbourhood intersecting only finitely many $f\_i^{-1}((0,1])$, say those indexed by $i\_0, i\_1, \ldots, i\_k$. Then the map
$$g\_x : V\_x \longrightarrow N\_k(\mathbb{R}, \leq) \times \Delta^k$$
given by $y \mapsto ((a\_{i\_0} \leq \cdots \leq a\_{i\_k}),(f\_{i\_0}(y), \ldots, f\_{i\_k}(y)))$ is clearly continuous, and hence so is the composition
$$V\_x \overset{g\_x}\longrightarrow N\_k(\mathbb{R}, \leq) \times \Delta^k \longrightarrow \vert N\_\bullet(\mathbb{R}, \leq) \vert.$$


I claim that this composition is equal to $g\vert\_{V\_x}$: as $f\_j(y)=0$ for $y \in V\_x$ and $j$ not one of $i\_0$, $i\_1$, ..., $i\_k$, due to how we chose $V\_x$, this is immediate. It follows that $g\vert\_{V\_x}$ is continuous; this holds for any $x$, so $g$ is continuous.


Edit: the addresses the elaboration in the edited question. For $x \in X$, let $V\_x$ be as above, intersecting the $f\_i^{-1}((0,1])$ indexed by $i\_0, i\_1, \ldots, i\_k$. For each $j$, if $x$ is not in $U\_{i\_j}$ then it is also not in the closure $\overline{f\_{i\_j}^{-1}((0,1])} \subset U\_{i\_j}$. Thus $V\_x$ may be replaced by the smaller open set $V\_x \cap (X \setminus \overline{f\_{i\_j}^{-1}((0,1])})$ which still contains $x$, but it does not intersect $f\_{i\_j}^{-1}((0,1])$ and so $U\_{i\_j}$ may be forgotten. After having discarded all such $U\_{i\_j}$, the set $V'\_x = V\_x \cap \bigcap\_{j=0}^k U\_{i\_j}$ is an open neighbourhood of $x$. 


Then the map
$$f\_x : V'\_x \longrightarrow N\_k(Y) \times \Delta^k$$
given by $y \mapsto (((a\_{i\_0},y) \leq \cdots \leq (a\_{i\_k},y)),(f\_{i\_0}(y), \ldots, f\_{i\_k}(y)))$ is well-defined and continuous. The remainder of the argument is identical.


