Let $X$ be a topological space and $X=U\cup V$ an open cover of $X$. Let $Z$ be the double mapping cylinder of the inclusions $U\leftarrow U\cap V\rightarrow V$. One has an obvious map $Z\rightarrow X$ by (identity on the top and the bottom of the cylinder and constant on the cylinder).
Is $Z\rightarrow X$ always a homotopy equivalence? If not, can on give mild conditions on $X$ or the open cover in which cases the map is a homotopy equivalence?
On
When the covering is numerable, i.e. it has a subordinate partition of unity (a numeration), then the obvious map $p$ from the double mapping cylinder to $X$ is a homotopy equivalence. The idea is that maps $X\to I$ give us a way of embedding $X$ into the quotient space of $U\sqcup U\cap V\times I \sqcup V$.
Let $J=\{U,V\}$, and let $N$ be the family of subsets $S$ of $J$ such that $U_S=\bigcap S$ is non-empty. Then $N$ is a simplicial complex. Whenever $T\subset S$ in $N$, let $D^T_S:\Delta_S\to\Delta_T$ denote the canonical linear embedding, where $\Delta_S$ is the standard-$n$-simplex for $n=|S|-1$, and let $r^S_T:U_S\hookrightarrow U_T$ be the inclusion of subspaces of $X$. We define $B$ as the quotient space
$$
B=\left(\coprod U_S\times \Delta_S\right)/\sim,\quad \text{with }\ (x,D^T_S(p))\sim(r^S_T(x),p)
$$
Each point in $B$ can be represented uniquely by a pair $(x,p)$, where $p=sU+tV$ is in the interior of a simplex $S$ and $x$ is an element of $U_S$.
Now let $C$ be the subspace of $X\times I^2$ consisting of the triples $y=(x,t_U,t_V)$ such that $t_u+t_v=1$ and $x\in \bigcap J(y)$ where $J(y)=\{W\in J\mid t_W>0\}$. There is then a continuous bijection $\rho:B\to C$ sending $(x,p)$ to $(x,t_U,t_V)$ such that $p=t_U U + t_V V$. This map need not be a homeomorphism, but one can show that it is a homotopy equivalence:
Let $\tau'_W=\max(0,t_W-1/3)$, and put $\tau_W=\tau'_W/(\tau'_U+\tau'_V)$. Then $(\tau_U,\tau_V)$ is a partition of unity subordinate to the open covering $\left(t_W^{-1}((0,1])\right)$ of $C$. Now define $$\pi:y=(x,t_U,t_V)\mapsto\left(x,\tau_U(y)+\tau_V(y)\right)=z$$ Let's show that $\pi$ is continuous. If $W\in J(z)$, then $0\ne t_W(z)=\tau_W(y)$, thus $t_W(y)\ne 0$, so $J(z)\subseteq J(y)$ and $x\in U_{J(y)}\subseteq U_{J(z)}$, hence the function $\pi$ is well-defined. To show continuity, let $w\in C$ be an arbitrary point. There is an open neighborhood $M$ of $w$ such that $J(M):=\{W\in J\mid M\cap\text{supp}(\tau_W)\ne\emptyset\}$ equals $\{W\in J\mid w\in\text{supp}(\tau_W)\}$, and $M$ is a subset of $t_W^{-1}((0,1])$ for every $W\in J(M)$. For any $y\in M$, we have $J(y)\supseteq J(M)$ and $x\in U_{J(M)}$, as well as $\pi(y)\in U_{J(M)}\times\Delta_{J(M)}$. That means that $\pi|_M$ is a continuous map $M\to U_{J(M)}\times\Delta_{J(M)}$, where the codomain has the product topology. One can show that the product topology is finer than the subspace topology of $U_{J(M)}\times\Delta_{J(M)}\subset B$, so $\pi|_M:M\to B$ is continuous. Since $M$ was a neighborhood of $w$, which was an arbitrary point of $C$, $\pi:C\to B$ is continuous.
Now a homotopy $\mathbf 1_B\simeq\pi\rho$ is given by $$\textstyle (y,s)=\left(x,\sum_W t_WW,s\right) \mapsto \left(x,\sum_W s\tau_W(\rho(y))+(1-s)t_W\right)$$ and a homotopy $\mathbf 1_C\simeq\rho\pi$ is given by $$(y,s)=(x,(t_W)_W,s) \mapsto (x,(s\tau_W(y)+(1-s)t_W)_W)$$
Now we can give sufficient conditions for the map $p:B\to X$ to be a homotopy equivalence.
Theorem: Let $(\tau_U,\tau_V)$ be a partition of unity subordinate to the cover $(U,V)$ of $X$. Then the map $p:B\to X$ is a homotopy equivalence.
Proof: Let $s:X\to C$ be the map $x\mapsto (x,(\tau_W(x))_W)$. Then we have $ps=\mathbf 1_X$. A homotopy $\mathbf 1_C\simeq sp$ is given by $$(x,(t_W)_W,s)\mapsto (x,(s\tau_W(x)+(1-s)t_W)_W)$$ Since $B$ is homotopy equivalent to $C$, we have $B\simeq X$.
I don't know.
The double mapping cylinder is $Z := ((U \cap V)\times [0,1] \sqcup U \sqcup V) / \sim $, where you identify $U \cap V \times \{0\}$ with its image via the inclusion into $U$, and $U \cap V \times \{ 1 \}$ with its image into $V$.
Now we describe the homotopy: first note that $U \cup V = U \sqcup V / \sim$, where we identify $U \cap V \subset U$ and $U \cap V \subset V$. But this is done with the double cone: let $x \in U \cap V$, then the line $x \times [0,1]$ connects $x \in U$ and $x \in V$. Collapsing all these lines at the same time gives you the homotopy from $Z$ to $U \cup V = X$. We still need the inverse construction.
If you want it more explicit, look at $U \sqcup V$ as $U \times \{0 \} \cup V \times \{1\}$, and the cylinder will consist of the points on $U \cap V \times \{t \}$, with $t \in [0,1]$. Here you can see that collapsing the lines of the cylinder is the same as identifying the points of $U \cap V$ downstairs and upstairs, and you obtain the desired homotopy. We still need the inverse construction.
EDIT: As Martha pointed out, we still need to construct a continuous map from $U \cup V$ to $Z$ which gives us an inverse modulo homotopy, and this is difficult in general. I can't think on any example now