The top answer in this question claims that it is possible to model the XOR gate using a 2-2-1 neural network (2 inputs, 2 neurons in the hidden layer) and no activation function. (Or an identity linear activation function if you prefer).
Is this true?
No.
A 2-2-1 linear neural network is a function
\begin{align} f: &\mathbb{R}^2 \rightarrow \mathbb{R}\\ f(x) &= B (A x + a) + b \qquad\text{with } x, a \in \mathbb{R}^{2}, A \in \mathbb{R}^{2 \times 2}, B \in \mathbb{R}^{1 \times 2}, b \in \mathbb{R} \end{align}
This is equivalent to
\begin{align} f(x) &= BAx + Ba + b \\ &= Cx + c \qquad\qquad \text{with } C \in \mathbb{R}^{1 \times 2}, c \in \mathbb{R} \end{align}
Obviously, $f$ is a linear function. Also quite obviously, the XOR-problem is not linearly seperable. Hence $f$ can't solve the XOR-problem.