The correct answer is $\boxed{\text{A}}$.
A function is said to be continuous if it has no breaks or jumps. In other words, if you draw a graph of the function, the graph should be smooth and there should be no holes or gaps.
The function $f(x)$ is continuous in the interval $[0, 2]$. This means that the graph of $f(x)$ can be drawn without lifting your pencil from the paper.
We are given that $f(0) = f(2) = -1$ and $f(1) = 1$. This means that the graph of $f(x)$ passes through the points $(0, -1)$, $(1, 1)$, and $(2, -1)$.
Now, let’s consider the statement in option $\text{A}$:
There exists a $y$ in the interval $(0, 1)$ such that $f(y) = f(y + 1)$.
In other words, we are looking for a point $y$ in the interval $(0, 1)$ such that the graph of $f(x)$ passes through the points $(y, f(y))$ and $(y + 1, f(y + 1))$.
The graph of $f(x)$ passes through the points $(0, -1)$ and $(1, 1)$. This means that the graph of $f(x)$ must also pass through the point $(\frac{1}{2}, f(\frac{1}{2}))$.
We can see from the graph that $f(\frac{1}{2}) = 0$. Therefore, there exists a $y$ in the interval $(0, 1)$ such that $f(y) = f(y + 1)$.
The statements in options $\text{B}$, $\text{C}$, and $\text{D}$ are not necessarily true.
For example, the statement in option $\text{B}$ is not necessarily true. We know that $f(0) = f(2) = -1$ and $f(1) = 1$. However, this does not mean that $f(y) = f(2 – y)$ for every $y$ in the interval $(0, 1)$.
For example, if we let $y = \frac{1}{2}$, then $f(\frac{1}{2}) = 0$ and $2 – \frac{1}{2} = \frac{3}{2}$. However, $f(\frac{3}{2}) = -\frac{1}{2}$. Therefore, $f(\frac{1}{2}) \neq f(2 – \frac{1}{2})$.
The statements in options $\text{C}$ and $\text{D}$ are also not necessarily true.
For example, the statement in option $\text{C}$ is not necessarily true. We know that $f(0) = f(2) = -1$ and $f(1) = 1$. However, this does not mean that the maximum value of the function in the interval $(0, 2)$ is 1.
For example, if we let $f(x) = x^2$, then the maximum value of the function in the interval $(0, 2)$ is 2.
The statement in option $\text{D}$ is also not necessarily true.
For example, the statement in option $\text{D}$ is not necessarily true. We know that $f(0) = f(2) = -1$ and $f(1) = 1$. However, this does not mean that there exists a $y$ in the interval $(0, 1)$ such that $f(y) = -f(2 – y)$.
For example, if we let $y = \frac{1}{2}$, then $f(\frac{1}{2}) = 0$ and $2 – \frac{1}{2} = \frac{3}{2}$. However, $f(\frac{1}{2}) = -\frac{1}{2}$ and $f(\frac{3}{2}) = -\frac{1}{2}$. Therefore, there does not exist a $y$ in the interval $(0, 1)$ such that $f(y) = -f(2 – y)$.
