elementary set theory - Notation on proving injectivity of a function $f:A^{B;cup; C}to A^Btimes A^C$

I'm trying to prove that for any cardinal numbers $a,b,c$, the following holds:
$a ^ {b + c} = a ^ b a ^ c$ i.e. that there exists a bijective function $f : A ^ {B \:\: \cup \:\: C} \rightarrow A^B \times A^C$

This is only part of the proof sketch I have (proving $f$ is injective), and I'd like to know if is well written, since I believe it has flaws.

---

Let $f: \{ g \:\:\: | g:B \cup C \rightarrow A \} \rightarrow \{ \langle g,h\rangle | \:\:\: g: B \rightarrow A \wedge h : C \rightarrow A \}$ such that

$f ( g_{b} \cup g_{c}) = \langle g_{b},g_{c}\rangle$.

Now,

$f( g_{b1} \cup g_{c1}) = f ( g_{b2} \cup g_{c2} ) \implies \langle g_{b1},g_{c1}\rangle = \langle g_{b2},g_{c2}\rangle$ and therefore $f$ is injective.

---

Questions:

1. Does that prove that $f$ is injective? I think it does not, since
   $f ( g_{b} \cup g_{c} ) = f ( g_{c} \cup g_{b}) \implies \langle g_{b},g_{c}\rangle = \langle g_{c},g_{b}\rangle (\bot)$


2. Is there an alternative way to define $f$? It's difficult for me to define it in terms of properties of elements of its domain.

Side note:

The title says "kinds" because the function domain and image sets are sets of sets, but I may be mistaken using that word, if so, please edit accordingly.

Answer

You can think of $g : B \cup C \to A$ as a couple of functions $g_B : B \to A$ and $g_C : C \to A$ such that for all $x \in B \cap C$, $g_B(x) = g_C(x)$. Therefore, letting $f : \Gamma \to \Delta$ (with

$$\Gamma = \{ g \,\, | \,\, g : B \cup C \to A \}, \qquad \Delta = \{ \langle g_B, g_C\rangle \, \, | \, \, g_B : B \to A, \, \, g_C : C \to A \}$$

obviously), you can see that a natural way to do this is to restrict the domain of $g : B \cup C \to A$ to just $B$ (or $C$), hence the map $f$ could naturally be defined by $g_B(x) = g(x)$ for all $x \in B$, $g_C(x) = g(x)$ for all $x \in C$ and $f(g) = \langle g_B, g_C \rangle$. In this manner the notation is more clear and more obviously well-defined.

This function $f$ is injective because of the following. Suppose that $f(g_1) = \langle g_B^1, g_C^1 \rangle = \langle g_B^2, g_C^2 \rangle =f(g_2)$. Thus for all $x \in B \cup C$, $x$ is either in $B$ or $C$, suppose $B$. Thus $g_1(x) = g_B^1(x) = g_B^2(x) = g_2(x)$. The case $x \in C$ is similar. Therefore $g_1 = g_2$.

I can't think of a natural way to define $F : \Delta \to \Gamma$ right now, it'll require some thinking, I guess. I am not familiar with the usual rigor abusive context, but my problem is that I don't want to assume $B \cap C = \varnothing$. If we can assume that there is an obvious way back (i.e. you can show that this $f$ is bijective, easily). If someone could comment on this question I'm asking, I'd love to read.