Bijections between disjoint sets Problem:Let $X,X'$ be disjoint sets and $Y,Y'$ be likewise disjoint. If $X\sim$ $Y$ and $X'\sim Y'$ then $X\cup X' \sim Y\cup Y'$ (for every sets $Z$ and $Z'$,the notation $Z \sim Z'$ means that there is a bijection from $Z$ to $Z'$).
My proof:
Let $f: X\rightarrow Y$ and $g: X'\rightarrow Y'$ be the respective bijections. Define $h: X\cup X' \rightarrow Y\cup Y'$ by $h(x)=$
$ \begin{cases} 
      f(x) & x\in X \\
      g(x) & x\in X' \\  
   \end{cases}$ 
. h is well defined:
Let $a=b$. Then either they are both in $X$ or both in $X'$, in either case, $f,g$ are well defined hence $h(a)=h(b)$.
Injective:
  Suppose $a\neq b$.
  Case (1): wlog suppose $a\in X$ and $b\in X'$. Then $f(a)\in Y$ and $g(b) \in Y'$, as $Y\cap Y' = \varnothing$ it follows that $f(a)\neq g(b)$ .
Case (2): wlog suppose $a,b \in X$ such that $a\neq b$. As $f$ is bijective, $f(a)\neq f(b)$ $\implies$ $h(a)\neq h(b)$.
Surjective:
  If $y\in Y\cup Y'$ then it is either in $Y$ or $Y'$, not both. As $f,g$ are surjective we can find an $x\in X'$ such that $f(x) = y$ . Similarly for $Y'$.
Can I have feedback, please?
 A: Yes, it is essentially correct. Just pay attention to the following things:


*

*In the proof of injectivity, there is a case (3): $a, b \in X'$ with $a \neq b$. Of course, this case is analogous to case (2), just replace $f$ with $g$ and use the injectivity of $g$ (maybe you omitted this case because it is similar to case (2), but it is better to mention it).

*In the proof of surjectivity, after you say that either $y \in Y$ or $y \in Y'$, not both, you should conclude that: in the first, case there is $x \in X$ (and not in $X'$, as you wrote instead, I guess it is a typo) such that $f(x) = y$, since $f$ is surjective; in the second case, there is $x \in X'$ such that $g(x) = y$, since $g$ is surjective. Therefore, in both cases there is $x \in X \cup X'$ such that $h(x) = y$.

*To be rigorous, your proof that $h$ is well defined is meaningless. Indeed, in this proof you use a functional notation for $h$ that makes sense only in the case you already know that $h$ is well defined. The rigorous approach would be to use first a relational notation for $h$, i.e. $h$ is defined as the binary relation from $X \cup X'$ to $Y \cup Y'$ such that 
\begin{align}
(x,y) \in h \text{ iff } 
\begin{cases} x \in X \text{ and } y = f(x), \text{ or } \\
x \in X' \text{ and } y = g(x)
\end{cases}
\end{align}
Now, proving that $h$ is a well defined function amounts to prove that: 
\begin{equation}\tag{*}
\text{for every } x \in X \cup X' \text{ there is exactly one } y \in Y \cup Y' \text{ such that } (x,y) \in h.
\end{equation} 
Only after proving that the functional notation $y = h(x)$ for $(x,y) \in h$ makes sense. Are you able to prove (*)?
