The nLab page on partitions 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.

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.

  • $\begingroup$ One example which seems to be in the general neighborhood of your query is the nerve theorem, as set out in Hatcher's Algebraic Topology math.cornell.edu/~hatcher/AT/AT.pdf (see section 4.G, starting at the page number 456, or 465 of the pdf file of the book). Does that help address your concerns? $\endgroup$ – Todd Trimble Mar 21 '15 at 18:26
  • $\begingroup$ You may prefer changing the definition of $g$ to say "where $i_0, \ldots, i_k$ is any set of indices that includes all $i$ where $f_i(x) \neq 0$". Of course, you have to check this is independent of the superset $\{i_0, \ldots, i_k\}$ chosen, but maybe the continuity is easier to see, since $\{x : f_{j}(x) = 0$ for all $j \notin \{i_0, \ldots, i_k\}\}$ is closed, while $\{x : f_{j}(x) = 0 \iff j \notin \{i_0, \ldots, i_k\}\}$ is the intersection of an open and a closed set. $\endgroup$ – Omar Antolín-Camarena Mar 23 '15 at 19:55
  • 1
    $\begingroup$ @OmarAntolín-Camarena That would be an option, but just shows me, that my example is bad chosen. In the situations I have in mind, the things corresponding to the $a_i$'s are only part of the nerve, if the corresponding function of the partition of unity is nonzero. Then one can not proceed as you suggested. $\endgroup$ – Tom Mar 24 '15 at 9:02
  • 2
    $\begingroup$ The proof of of (4.1) in Segal's "Classifying Spaces and Spectral Sequences" (math.northwestern.edu/~konter/gtrs/segal.pdf) should exemplify my doubts more closely. I don't see the map $\phi\colon X\rightarrow BR_U$ being continuous. Do you or somebody else do? $\endgroup$ – Tom Mar 24 '15 at 9:05
  • $\begingroup$ Note there is a typo in the statement of Segal's Proposition 4.1: it should say that $\text{pr}\colon BX_U\to X$ is a homotopy equivalence, not $BR_U\to X$. $\endgroup$ – Tom Church Apr 29 '15 at 0:20

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.

  • $\begingroup$ This goes along with the comment of @Omar Antolín-Camarena, settles the example as posed, but does not address my concerns, since my example was chosen badly. I edited the question with an extended example, which should exemplify my uncertainty more closely. $\endgroup$ – Tom Apr 28 '15 at 12:16
  • $\begingroup$ Why does $x\in U_{i_j}$ hold? It think it only holds $U_{i_j}\cap V_x\neq \emptyset$, but this does not imply $x\in U_{i_j}$. If I would choose only the subset of indices $\{i'_0,...,i'_{k'}\}\subseteq\{i_0,...,i_k\}$ where $x\in U_{i'_j}$ holds to fix this then the $f_{i'_j}$ maybe do not sum up to $1$ on $V_x\bigcap_{j=0}^{k'}U_{i'_{k'}}$. $\endgroup$ – Tom May 1 '15 at 15:31
  • $\begingroup$ You are quite right: I have edited my answer to show how these $U_{i_j}$'s can be eliminated. $\endgroup$ – Oscar Randal-Williams May 1 '15 at 16:17
  • $\begingroup$ That settles the example. Thanks. It even shows that the map $g$ factors over $|N_\bullet(Y')|$, where $Y'$ is the same simplicial space as $Y$ but endowing $\mathbb{R}$ with the discrete topology. $\endgroup$ – Tom May 2 '15 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.