Injectivity and surjectivity of $f\colon P(X) \times P(Y) \to P(X \times Y)$ Let $X$ and $Y$ be nonempty sets and a function such as:

$$f: P(X) \times P(Y) \to P(X \times Y)$$
$$(A,B)\to A \times B$$

The question is if this function is injective or surjective, but I do not even understand the function. If $X={1,2,3}$ and $Y={1,2}$, $P(X)=\{\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\},\{1,2,3\},0\}$ and $P(Y)=\{\{1\},\{2\},\{1,2\},0\}$, who is $f(\{1,2\},\{1,2,3\})$?
First, to understand the function itself:

$f:\mathcal{P}(X)\times\mathcal{P}(Y)\to \mathcal{P}(X\times Y)~~~~~(A,B)\mapsto A\times B$

The function takes an ordered pair of sets, $(A,B)$ where $A$ is a subset of $X$ and $B$ is a subset of $Y$, and maps it to the set of all ordered pairs of the elements where the first element is in $A$ and the second is in $B$.

Your specific example, letting $X=\{1,2\}$ and $Y=\{1,2,3\}$ we have

$f(\color{red}{\{1,2\}},\color{blue}{\{1,2,3\}}) = \{(\color{red}{1},\color{blue}{1}),(\color{red}{1},\color{blue}{2}),(\color{red}{1},\color{blue}{3}),(\color{red}{2},\color{blue}{1}),(\color{red}{2},\color{blue}{2}),(\color{red}{2},\color{blue}{3})\}$

Similarly:

$f(\color{red}{\{2\}},\color{blue}{\{1,3\}})=\{(\color{red}{2},\color{blue}{1}),(\color{red}{2},\color{blue}{3})\}$



As for the question of injectivity, assuming that $X$ is nonempty and contains at least one element $x$:

$f(\emptyset,\emptyset)=\emptyset\times \emptyset=\emptyset$

also

$f(\{x\},\emptyset)=\{x\}\times\emptyset = \emptyset$

So, we have shown that there exist two inputs, $(\emptyset,\emptyset)$ and $(\{x\},\emptyset)$ which both map to the same output, therefore so long as $X$ is nonempty the function will not be injective.  Reversing the roles of $X$ and $Y$ will show the same result for if $Y$ is nonempty.  We get as a result, $f$ is injective if and only if $X$ and $Y$ are simultaneously empty and is not injective otherwise.



As for the question of surjectivity.  You should know 

$|\mathcal{P}(X)\times \mathcal{P}(Y)|=|\mathcal{P}(X)|\times|\mathcal{P}(Y)|=2^{|X|}\times 2^{|Y|}=2^{|X|+|Y|}$

and

$|\mathcal{P}(X\times Y)|=2^{|X\times Y|} = 2^{|X|\times |Y|}$

For large enough sets $X$ and $Y$, one has $2^{|X|+|Y|}&lt;2^{|X|\times|Y|}$ which proves that $f$ cannot be surjective.  If $X$ and/or $Y$ are small enough however this isn't true and isn't sufficient to prove or disprove surjectivity.  For example if $X$ or $Y$ are empty, then the function will indeed be surjective.

For an explicit example, note that that with $X=\{1,2\}$ and $Y=\{1,2,3\}$ there is no input which will map to $\{(1,1),(2,1),(2,2)\}$ despite this being an element of the codomain.  All of the outputs of the function will be "balanced" so to speak.
$$f(\{1,2\},\{1,2,3\}) = \{1,2\} \times \{1,2,3\} = \{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)\}$$