Let f be an injective map with domain {x, y, z} and range {1, 2, 3},

Solution:

Case-1: Let $f(x)=1$ be true.

Then, $f(y) \neq 1$ and $f(z) \neq 2$ are false.

So, $f(y)=1$ and $f(z)=2$

$\Rightarrow f(x)=1, f(y)=1$

$\Rightarrow x$ and $y$ have the same images.

This contradicts the fact that $f$ is one-one.

Case-2: Let $f(y) \neq 1$ be true.

Then, $f(x)=1$ and $f(z) \neq 2$ are false.

So, $f(x) \neq 1$ and $f(z)=2$

$\Rightarrow f(x) \neq 1, f(y) \neq 1$ and $f(z)=2$

$\Rightarrow$ There is no pre-image for $1 .$

This contradicts the fact that range is $\{1,2,3\}$.

Case-3: Let $f(z) \neq 2$ be true.

Then, $f(x)=1$ and $f(y) \neq 1$ are false.

So, $f(x) \neq 1$ and $f(y)=1$

$\Rightarrow f(x)=2, f(y)=1$ and $f(z)=3$

$\Rightarrow f(y)=1$

$\Rightarrow f^{-1}(1)=y$

So, the answer is (b).