[proofplan] We prove first the finite open good-cover case. For an open cover in a Euclidean subspace, a distance-to-the-complement construction gives a finite [partition of unity](/page/Partition%20of%20Unity) whose barycentric map lands in the geometric realization of the nerve. The homotopy equivalence is then supplied by the standard finite [Nerve Theorem](/theorems/6468), in the form for finite open covers of paracompact Hausdorff spaces with contractible nonempty finite intersections. Open convex covers satisfy these hypotheses because finite intersections of convex sets are convex and hence contractible when nonempty. In the model case, the model space is assumed paracompact Hausdorff, so the same finite Nerve Theorem applies directly to the model cover before the homotopy type and intersection pattern are transferred back. [/proofplan]

[step:Construct barycentric coordinates supported on the cover]Assume first that $\mathcal F=\{C_i\}_{i\in I}$ is a finite open cover of its union and that every nonempty finite intersection is contractible. Define the union \begin{align*} X:=\bigcup_{i\in I} C_i\subset \mathbb R^d. \end{align*} Since each $C_i$ is open in $X$, define for each $i\in I$ the function \begin{align*} \rho_i:X&\to [0,1] \end{align*} by setting $\rho_i(x):=1$ if $X\setminus C_i=\varnothing$, and otherwise \begin{align*} \rho_i(x):=\min\{1,\operatorname{dist}(x,X\setminus C_i)\}, \end{align*} where the distance is computed using the Euclidean metric restricted to $X$. The map $x\mapsto \operatorname{dist}(x,X\setminus C_i)$ is $1$-Lipschitz on $X$, so $\rho_i$ is continuous. Also, if $x\in C_i$, then the openness of $C_i$ in $X$ gives a number $\varepsilon>0$ such that \begin{align*} \{y\in X: |y-x|<\varepsilon\}\subset C_i. \end{align*} Hence every point of $X\setminus C_i$ has Euclidean distance at least $\varepsilon$ from $x$, so $\operatorname{dist}(x,X\setminus C_i)\geq \varepsilon$ when $X\setminus C_i\neq\varnothing$, and $\rho_i(x)>0$ in all cases. Conversely, if $\rho_i(x)>0$, then $x\notin X\setminus C_i$, hence $x\in C_i$. Define the finite sum \begin{align*} \rho:X&\to (0,\infty) \end{align*} by \begin{align*} \rho(x):=\sum_{i\in I}\rho_i(x). \end{align*} For each $x\in X$, some $i\in I$ satisfies $x\in C_i$, and the preceding paragraph gives $\rho_i(x)>0$; hence $\rho(x)>0$. For each $i\in I$, define \begin{align*} \varphi_i:X&\to [0,1] \end{align*} by \begin{align*} \varphi_i(x):=\frac{\rho_i(x)}{\rho(x)}. \end{align*} Then each $\varphi_i$ is continuous, $\sum_{i\in I}\varphi_i(x)=1$ for every $x\in X$, and $\varphi_i(x)>0$ implies $x\in C_i$. Let $e_i$ denote the vertex indexed by $i$ in the standard simplex on $I$. Define \begin{align*} F:X&\to |\mathcal N(\mathcal F)| \end{align*} by \begin{align*} F(x):=\sum_{i\in I}\varphi_i(x)e_i. \end{align*} For $x\in X$, define the support set \begin{align*} S(x):=\{i\in I:\varphi_i(x)>0\}. \end{align*} If $i\in S(x)$, then $x\in C_i$, so \begin{align*} x\in \bigcap_{i\in S(x)}C_i. \end{align*} Thus $S(x)$ is a simplex of $\mathcal N(\mathcal F)$, and $F(x)$ lies in the corresponding geometric simplex. Therefore $F$ is a well-defined continuous map from $X$ to $|\mathcal N(\mathcal F)|$.[/step]

[guided]We first build the canonical barycentric map associated to the cover. Define \begin{align*} X:=\bigcup_{i\in I} C_i\subset \mathbb R^d. \end{align*} The point of the construction is to assign to each $x\in X$ barycentric coordinates indexed only by those cover elements that contain $x$. For each $i\in I$, define a function \begin{align*} \rho_i:X&\to [0,1] \end{align*} as follows. If $X\setminus C_i=\varnothing$, set $\rho_i(x):=1$ for all $x\in X$. If $X\setminus C_i\neq\varnothing$, set \begin{align*} \rho_i(x):=\min\{1,\operatorname{dist}(x,X\setminus C_i)\}, \end{align*} where the distance is taken in the Euclidean metric restricted to $X$. The distance-to-a-set function is $1$-Lipschitz on a [metric space](/page/Metric%20Space), so $x\mapsto \operatorname{dist}(x,X\setminus C_i)$ is continuous on $X$. The map $t\mapsto \min\{1,t\}$ is continuous on $[0,\infty)$, and the constant function $1$ is continuous. Therefore each $\rho_i$ is continuous. We now verify the support property carefully. If $x\in C_i$, then because $C_i$ is open in $X$, there exists $\varepsilon>0$ such that \begin{align*} \{y\in X: |y-x|<\varepsilon\}\subset C_i. \end{align*} Consequently no point of $X\setminus C_i$ lies within Euclidean distance $\varepsilon$ of $x$. If $X\setminus C_i\neq\varnothing$, this gives \begin{align*} \operatorname{dist}(x,X\setminus C_i)\geq \varepsilon, \end{align*} and hence $\rho_i(x)>0$. If $X\setminus C_i=\varnothing$, then $\rho_i(x)=1>0$. Conversely, if $\rho_i(x)>0$, then $x$ cannot belong to $X\setminus C_i$, since a point has distance zero from any set containing it. Thus $\rho_i(x)>0$ implies $x\in C_i$. Define \begin{align*} \rho:X&\to (0,\infty) \end{align*} by \begin{align*} \rho(x):=\sum_{i\in I}\rho_i(x). \end{align*} The sum is finite because $I$ is finite. It is strictly positive because every $x\in X$ lies in at least one member $C_i$ of the cover, and for such an index the preceding paragraph gives $\rho_i(x)>0$. For each $i\in I$, define \begin{align*} \varphi_i:X&\to [0,1] \end{align*} by \begin{align*} \varphi_i(x):=\frac{\rho_i(x)}{\rho(x)}. \end{align*} The denominator is positive everywhere, so this quotient is well-defined and continuous. Since $\rho$ is the sum of the $\rho_i$, we have \begin{align*} \sum_{i\in I}\varphi_i(x)=1 \end{align*} for every $x\in X$. Also, $\varphi_i(x)>0$ is equivalent to $\rho_i(x)>0$, and hence implies $x\in C_i$. Let $e_i$ denote the vertex indexed by $i$ in the standard simplex on $I$. Define \begin{align*} F:X&\to |\mathcal N(\mathcal F)| \end{align*} by \begin{align*} F(x):=\sum_{i\in I}\varphi_i(x)e_i. \end{align*} To check that this lands in the nerve, define \begin{align*} S(x):=\{i\in I:\varphi_i(x)>0\}. \end{align*} For every $i\in S(x)$, the support property gives $x\in C_i$. Hence \begin{align*} x\in \bigcap_{i\in S(x)}C_i. \end{align*} This intersection is nonempty, so $S(x)$ is a simplex of $\mathcal N(\mathcal F)$. Therefore the barycentric point $F(x)$ lies in the geometric simplex indexed by $S(x)$, and $F:X\to |\mathcal N(\mathcal F)|$ is well-defined and continuous.[/guided]

[step:Apply the finite Nerve Theorem to the open good cover]We use the following standard external theorem: the finite Nerve Theorem states that if $X$ is a paracompact [Hausdorff space](/page/Hausdorff%20Space) and $\{V_i\}_{i\in I}$ is a finite open cover of $X$ such that every nonempty finite intersection $\bigcap_{j\in J}V_j$ is contractible, then $X$ is homotopy equivalent to $|\mathcal N(\{V_i\}_{i\in I})|$. In the open-cover case under consideration, $X\subset\mathbb R^d$ is open as a finite union of open subsets of $\mathbb R^d$. Hence $X$ is Hausdorff and paracompact. The family $\{C_i\}_{i\in I}$ is finite, each $C_i$ is open in $X$, and every nonempty finite intersection is contractible by hypothesis. Therefore the finite Nerve Theorem applies and gives \begin{align*} X\simeq |\mathcal N(\mathcal F)|. \end{align*} Since $X=\bigcup_{i\in I}C_i$, this proves the finite open good-cover case. If each $C_i$ is open and convex in $\mathbb R^d$, then every nonempty finite intersection \begin{align*} \bigcap_{j\in J}C_j \end{align*} is open and convex. A nonempty convex subset of $\mathbb R^d$ is contractible by the straight-line homotopy to any chosen point in it. Thus the open convex case satisfies the finite open good-cover hypotheses, and the conclusion follows.[/step]

[guided]The barycentric map identifies the correct combinatorial target, but the homotopy equivalence itself is supplied by the finite Nerve Theorem. We use the following precise form: if $X$ is a paracompact Hausdorff space and $\{V_i\}_{i\in I}$ is a finite open cover of $X$ such that every nonempty finite intersection $\bigcap_{j\in J}V_j$ is contractible, then \begin{align*} X\simeq |\mathcal N(\{V_i\}_{i\in I})|. \end{align*} We verify the hypotheses for the finite open good-cover case. The space is \begin{align*} X:=\bigcup_{i\in I}C_i\subset\mathbb R^d. \end{align*} Because the $C_i$ are open in $\mathbb R^d$ in the open-cover case, $X$ is open in $\mathbb R^d$. Every subspace of $\mathbb R^d$ is Hausdorff, and every open subspace of the metric space $\mathbb R^d$ is paracompact. The cover is finite because $I$ is finite. Each $C_i$ is open in $X$ because it is open in $\mathbb R^d$ and contained in $X$. Finally, the nonempty finite intersections are contractible by the good-cover hypothesis. The finite Nerve Theorem therefore gives \begin{align*} X\simeq |\mathcal N(\mathcal F)|. \end{align*} Since $X$ is exactly $\bigcup_{i\in I}C_i$, this proves the open good-cover case. Now suppose each $C_i$ is open and convex. For every nonempty finite subset $J\subseteq I$, the intersection \begin{align*} C_J:=\bigcap_{j\in J}C_j \end{align*} is convex because intersections of convex sets are convex. If $C_J$ is nonempty, choose a point $p_J\in C_J$ and define \begin{align*} H_J:C_J\times [0,1]&\to C_J \end{align*} by \begin{align*} H_J(x,t):=(1-t)x+t p_J. \end{align*} Convexity guarantees $H_J(x,t)\in C_J$ for all $x\in C_J$ and $t\in[0,1]$. This is a contraction of $C_J$ to $p_J$, so $C_J$ is contractible. Hence an open convex cover is a finite open good cover, and the conclusion follows from the preceding paragraph.[/guided]

[step:Transfer the conclusion through the good-cover model] Assume now that $\mathcal F=\{C_i\}_{i\in I}$ admits a finite open good-cover model $\mathcal U=\{U_i\}_{i\in I}$ as in the statement. Define \begin{align*} Y:=\bigcup_{i\in I}U_i. \end{align*} By the model hypothesis, $Y$ is paracompact Hausdorff, $\mathcal U$ is a finite open cover of $Y$, and every nonempty finite intersection of members of $\mathcal U$ is contractible. Applying the finite Nerve Theorem to the cover $\mathcal U$ of $Y$ gives \begin{align*} Y\simeq |\mathcal N(\mathcal U)|. \end{align*} For every nonempty finite subset $J\subseteq I$, the model preserves the finite intersection pattern: \begin{align*} \bigcap_{j\in J}C_j\neq\varnothing \end{align*} if and only if \begin{align*} \bigcap_{j\in J}U_j\neq\varnothing. \end{align*} Therefore the identity map on the vertex set $I$ induces an isomorphism of simplicial complexes \begin{align*} \mathcal N(\mathcal F)\cong \mathcal N(\mathcal U), \end{align*} and hence a homeomorphism of geometric realizations \begin{align*} |\mathcal N(\mathcal F)|\cong |\mathcal N(\mathcal U)|. \end{align*} The model also supplies a homotopy equivalence \begin{align*} \bigcup_{i\in I}C_i\simeq Y. \end{align*} Combining these equivalences yields \begin{align*} \bigcup_{i\in I}C_i\simeq Y\simeq |\mathcal N(\mathcal U)|\cong |\mathcal N(\mathcal F)|. \end{align*} Thus $\bigcup_{i\in I}C_i$ has the homotopy type of $|\mathcal N(\mathcal F)|$. [/step]