Computable Injective Function

Let's analyse each statement one by one. $1.\text{ If} f(x) = 1 \text{ is true}$ Then, $f(y) \neq 1$ is also true.But only one statement is true. So $f(x) =1$ is false statement. $2. \text{If} f(y) \neq 1 \text{is true}$ Then, $f(x) = 1$ and $f(z) \neq 2$ is false. So $f(x) \neq 1$ and also $f(y) \neq 1$ so, $f(z)= 1$ . Hence, $f(z) \neq 2$ is also true. But only one of given statement is true. So $f(y) \neq 1$ is false statement. $3. \text{ If} f(z) \neq 2 \text{is true}$ Then, $f(x) = 1$ and $f(y) \neq 1$ are false. So $f(x)\neq 1$ and $f(y) =1$ . Also we can conclude $f(z) = 3 \text{and} f(x) = 2$ . Hence , $\boxed{f^{-1}(1) = y}$ .

1. $f(x) = 1$
2. $f(y) \neq 1$
3. $f(z) \neq 2$

If $f^{-1}(1) = x$ then 1 and 2 are true. But only 1 of them is true, then $f^{-1}(1) \neq x$ .

If $f^{-1}(1) = z$ then 3 and 2 are true. Then $f^{-1}(1) \neq z$ .

We can state that $y = 1; x = 2; z = 3$ . Then 1 and 2 are false and 3 is true.

Remark, all values in the range must be mapped to by a unique value in the domain (unique here, since f is 1-1), since otherwise, any such value that is not mapped to does not belong in the range.

Now,

Statement 1. f(x) = 1

must be false since it immediately follows that statement 2. f(y) != 1

must be true as well, but we are only allowed a single true statement.

Let us look at statement 2 next.

1. f(y) != 1

If statement 2 is indeed true, then it follows that either f(y) = 2 or f(y) = 3

If f(y) = 2, it immediately follows that statement 3 is true as well; again contradicting the assumption that there's only a single true statement in our collection of assertions.

If f(y) = 3, then f(x) must equal 2, since we assume statement 1. is false, as is (i.e. f(x) != 1 and therefore f(x) = 2 since that's the only value in the range set that is left [remember, f is 1-1]). But f(x) = 2 implies statement 3. f(z) != 2 is true as well, which is a contradiction yet again.

I.e., the assumption that statement 2 is true, yields a contradiction. Therefore, this assumption must be false. It follows that statement 2 is false, and f(y) = 1.

Certainly, f^{-1}(1) = y, as desired.