Interpretation 1: Given the data in the problem, I think you're interested in a pure-exchange economy with externalities i.e.,
- $u_1(x_1,y_2) = x_1^\beta y_2^{1-\beta}$, $u_2(x_2,y_2) = \min(x_2,y_2)$, where $\beta\in (0,1)$
- $\omega_1=(0,1)$ and $\omega_2=(1,0)$
Set of feasible allocations is given by
$\mathcal{F} = \{((x_1,y_1),(x_2,y_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1+x_2=1 \ \wedge \ y_1+y_2=1\}$
Proposition 0: Set of Pareto efficient allocations is non-empty.
Proof: Consider $((x_1,y_1),(x_2,y_2)) = ((0,0),(1,1))$. This allocation is Pareto efficient because if we move to any other feasible allocation, that will necessarily make individual $2$ worse off.
Proposition 1: Set of Pareto efficient allocations is given by the set $\{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_2=1\}$
Proof: Consider a feasible allocation $((x_1,y_1),(x_2,y_2))$ and suppose $y_2 < 1$ holds. Clearly, $((x_1,0),(x_2,1))$ is feasible and is a Pareto improvement over $((x_1,y_1),(x_2,y_2))$ as $1$ will be made better off without making $2$ worst off. Therefore, $y_2=1$ is a necessary condition for efficiency. If we consider the set $\{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_2=1\}$, it yields the utility possibilities as $u_1^{1/\beta}+u_2=1$, where $0 \leq u_1 \leq 1$, which is a strictly decreasing curve in the $(u_1,u_2)-$space establishing that $y_2=1$ along with feasibility yields the set of efficient allocations.
Interpretation 2 (If you made a typo in writing utilities): Given a pure-exchange economy,
- $u_1(x_1,y_1) = x_1^\beta y_1^{1-\beta}$, $u_2(x_2,y_2) = \min(x_2,y_2)$, where $\beta\in (0,1)$
- $\omega_1=(0,1)$ and $\omega_2=(1,0)$
Set of feasible allocations is given by
$\mathcal{F} = \{((x_1,y_1),(x_2,y_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1+x_2=1 \ \wedge \ y_1+y_2=1\}$
Proposition 0: Set of Pareto efficient allocations is non-empty.
Proof: Consider $((x_1,y_1),(x_2,y_2)) = ((1,1),(0,0))$. This allocation is Pareto efficient because if we move to any other feasible allocation, that will necessarily make individual $1$ worse off.
Proposition 1: If a feasible allocation $((x_1,y_1),(x_2,y_2))$ satisfy $x_2\neq y_2$, then it is not Pareto efficient.
Proof: Consider a feasible allocation $((x_1,y_1),(x_2,y_2))$ and suppose $x_2 < y_2 \leq 1$ holds. Therefore, $x_1 = 1-x_2 > 0$. Clearly, $((x_1,y_1+y_2-x_2),(x_2,x_2))$ is feasible and Pareto Superior to $((x_1,y_1),(x_2,y_2))$. Therefore, $((x_1,y_1),(x_2,y_2))$ is not Pareto efficient. Likewise, by a symmetric argument, a feasible allocation $((x_1,y_1),(x_2,y_2))$ satisfying $y_2 < x_2 \leq 1$ is also not Pareto efficient.
Equivalently, in proposition 1, we have shown that if an allocation is Pareto efficient then it satisfy the condition $x_2 =y_2$. Now, we'll show that the converse is also true.
Proposition 2: Any feasible allocation $((x_1,y_1),(x_2,y_2))$ satisfying $x_2 = y_2$ is Pareto efficient.
Proof: Observe that the sum of the utilities of the two individuals at all the feasible allocations satisfying $x_2= y_2$ equals $u_1+u_2=x_1^\beta y_1^{1-\beta} + \min(x_2,y_2) = x_1^\beta x_1^{1-\beta} + x_2 = 1$, and is maximum among all feasible allocations (using the argument in proposition 1), therefore it follows that all the allocations satisfying $x_2 = y_2$ are Pareto efficient.