If $A$ and $B$ have $n$ and $m$ elements respectively with $A\cap B=\emptyset$. Prove that $A\cup B$ has $m+n$ elements. If $A$ and $B$ have $n$ and $m$ elements respectively with $A \cap B=\emptyset$. Prove that $A \cup B$ has $m+n$ elements.
The solution in the book starts by letting $f$ be a bijection from $\{$1,..m$\}$ onto $A$ and $g$ a bijection from $\{$1,..n$\}$ onto $B$
Then proving the function
$h:=$$f(i)$ if $i=1,...,m$
Or $g(i-m)$ if $i=m+1,...,m+n$ is a bijection from $\{$1,..,m,m+1,..,m+n$\}$ onto $A$$\cup$$B$
But I'm stuck with the surjection part

Follows immediately from the result here
Proving that for finite sets $m(A \cup B)=m(A)+m(b)-m(a \cap B)$; what is a good level of rigor?
that
$m(A \cup B)=m(A)+m(B)-m(A \cap B)$.
