Proof that function is bijection I have a problem to prove that $f(a,b)=(g(a),h(b))$ function is bijection, given $g$ is bijection and $h$ is bijection.
I approached a problem as to prove that it is surjective and injective, which led me to proofs by contradiction of those, which in its turn to negation of first order logic statements for definitions of surjective and injective functions. And more and more... 
After writing about a page of such proof I asked instructor to review it as it was to complex for such as task, just to look at. He advised, that it is to complex and really first order logic are not so required for the proof.
As alternative solution I asked if it is ok if I just say $f(a,b)=(g(a), h(b))$ is a composition of two bijections:
$$u(a)=(g(a), b)$$
and:
$$v(b)=(c, h(b))$$
and since $f=uºv$, and $u$ and $v$ are bijections, then $f$ is bijection. 
I was then advised to prove that $u$ and $v$ are bijections and composition of function also needs proof.
Sorry for long introduction. Can anyone point me to similar nicely written proof. Or a book. I just need to understand style. 
Thanks you very much,
 A: To prove that the two are surjections prove that for every $(c,d)$ in codomain $\exists (a,b)$ in domain s.t. $(g(a),h(b)) = (c,d)$. But from the surjectivity of $h$ there exists $a$, s.t. $g(a) = c$ and from the surjectivity of $g$ there exists $b$, s.t. $h(b) = d$. So therefore there exists $(a,b)$ in domain s.t. $(g(a),h(b)) = (c,d)$. Hence $f$ is surjective.
Now assume that $f(a_1,b_1) = f(a_2,b_2) \iff (g(a_1),h(b_1)) = (g(a_2),h(b_2)) \iff g(a_1) = g(a_2)$ and $h(b_1) = h(b_2) \iff a_1 = a_2$ and $b_1 =b_2 \iff (a_1,b_1) = (a_2,b_2)$. Hence $f$ is injective.
Finally combining these two we can conclude that $f$ is bijective.
A: I'd just do a straightforward (non-contradiction) proof of surjective and injective.
To prove $f$ is injective, let $f(a_1,b_1) = f(a_2,b_2)$. Then by definition of $f$, $$(g(a_1),h(b_1)) = (g(a_2),h(b_2))$$
which implies $g(a_1) = g(a_2)$ and $h(b_1) = h(b_2)$. Now since $g$ and $h$ are injective, we have $a_1 = a_2$ and $b_1 = b_2$, which proves $f$ is injective.
The proof that $f$ is surjective is similarly straightforward. However, you need to be careful about what your codomain is. In particular, if the codomain of $f$ is larger than $\text{Im}(g) \times \text{Im}(h)$, then $f$ will not be surjective.
