Function theory and set theory with cartesian products Let $S$ and $T$ be sets and define the function
$$f:\mathcal P(S) \times \mathcal P (T)\to \mathcal P(S \cup T)$$
by $f(A,B) = A \cup B$ for all $A \subseteq S$ and all $B \subseteq T$. Prove that $f$ is a surjection
My problem is $f$ doesn't seem like a surjection.  if $A$ and $B$ are subsets of $S$ and $T$ then it seems entirely possible that $A \cup B$ doesn't span the entire codomain.  And if it does, do you just use english to explain it? 
 A: Look at the case $S = \{1, 2\}$, $T = \{ Q \}$. Your concern (I think) is that $H = \{1, Q\}$ doesn't seem as if it'd be in the image of $f$. But look at 
$$
A = H \cap S = \{ 1 \} \\
B = H \cap T = \{ Q \}
$$
Clearly $A$ is a subset of $S$ and $B$ a subset of $T$, and $f(A, B) = A \cup B = H$. 
And this idea works in general: take your target set $U$ in the codomain, and intersect it with $S$ and $T$ to get sets $A$ and $B$ with $f(A, B) = U$. 
The shortest form of this answer is thus:
$$
f(U \cap S, U \cap T) = (U \cap S) \cup (U \cap T) = U \cap (S \cup T) = U 
$$
for any $U \subset S \cup T$. 
A: What is a subset of $S\cup T$?  It is a collection of elements, some of them from $S$, some of them from $T$.  Let $A$ the subset of these elements that are from $S$.  Let $B$ be the subset of these elements from $T$.  Then $A\cup B$ is the subset and $A\subseteq S$ and $B\subseteq T$.
A: The statement only says that every subset of $S\cup T$ is the union of a subset of $S$ and a subset of $T$.
If $A\subseteq S\cup T$ then $A\cap S\subseteq S$ and $A\setminus S\subseteq T$ and that does it.  So does $A\setminus T$ and $A\cap T$, and those are two different ways of doing it if $A\cap S\cap T\ne\varnothing$.
