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Let $(U_i\subset U)$ be a family of subspaces of a topological space $U$. Consider the Čech nerve of this family, given by the simplicial object furnished by taking iterated kernel pairs of the associated arrow $\amalg _iU_i\to U$. The colimit of this diagram seems to be a reasonable candidate for the union of the $U_i$ in $U$, and so more generally for the "union of the images" for a general family $(U_i\to U)$.

For what families of subspaces $(U_i\subset U)$ is the colimit of the Čech nerve realized by the union? For instance, is it true if the $U_i$ are all open? Is it true if the $U_i$ merely contain an open subspace of $U$?

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First of all, there's no need to consider the entire Čech nerve, since just the last two terms $\coprod_{i,j} U_i\cap U_j \rightrightarrows\coprod_i U_i$ are cofinal in the entire diagram. We also may as well assume $\bigcup U_i$ is all of $U$ (if not, replace $U$ with $\bigcup U_i$). The colimit is then just the quotient space of $\coprod U_i$ given by the natural map $\coprod U_i\to U$. That is, the colimit has underlying set $U$, and a set $V\subseteq U$ is open in the colimit iff $V\cap U_i$ is open in $U_i$ for each $i$.

There isn't any simple characterization of when this topology is the original topology on $U$, but there are many sufficient conditions. For instance, it suffices for each $U_i$ to be open, or in fact just for the interiors of the $U_i$ to cover $U$. Indeed, in that case, if $V\cap U_i$ is open in $U_i$ for each $i$, that means for each $x\in V$, $V$ contains a neighborhood of $x$ in $U$ (consider $U_i$ such that $x$ is in the interior of $U_i$), and thus is open.

For an example of what can go wrong if the interiors of the $U_i$ do not cover $U$ (even if they are dense in $U$), let $U$ be a closed disk and for each $x$ in the boundary of the disk, let $U_x$ be the open disk together with $x$. Since the open disk is dense in $U$, the interiors of the $U_x$ are dense in $U$. However, the colimit topology is not the same as the original topology on $U$. Indeed, in the colimit topology, every subset of the boundary of the disk is closed, since its intersection with each $U_x$ has at most one point and hence is closed.

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  • $\begingroup$ Is there a slick way of proving the cofinality of the last two terms? $\endgroup$ – Arrow Oct 9 '17 at 9:07
  • $\begingroup$ Well, in this example at least, it's obvious that the last two terms have the same colimit as the entire diagram. The colimit is the last term modulo the equivalence relation given by all parallel pairs of arrows to it from other objects. A pair of such parallel arrows just identify the copies of $U_{i_1}\cap\dots\cap U_{i_n}$ sitting inside $U_{i_j}$ and $U_{i_k}$ for some $j$ and $k$. But these identifications are all just a subset of the identifications given by the pairwise intersections. $\endgroup$ – Eric Wofsey Oct 9 '17 at 15:58
  • $\begingroup$ It's also quite easy to verify the definition of cofinality once you write down what it means. You have the category $\Delta$ of finite nonempty sets and the subcategory $I$ of sets of cardinality 1 and 2. The claim is that for any finite nonempty set $s$, the comma category $(I\downarrow s)$ has exactly one connected component. Obviously it is nonempty, since any element of $s$ gives a map $\{*\}\to s$. And given two maps $\{*\}\to s$, they can be factored through a single map from a set with two elements to $s$, so they are in the same connected component of $(I\downarrow s)$. $\endgroup$ – Eric Wofsey Oct 9 '17 at 16:05
  • $\begingroup$ And every map from a set with two elements to $s$ can be restricted to a set with one element. So every object of $(I\downarrow s)$ is in the same connected component as a map $\{*\}\to s$, and all such maps are in a single connected component, so $(I\downarrow s)$ is connected. $\endgroup$ – Eric Wofsey Oct 9 '17 at 16:06

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