For which families of subsets is the colimit of the Čech nerve realized by the union? 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 $U$?
 A: 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.
