Is this function bijective? I am studying probability, and need to use some transformation results quite a lot. The only issue I have with these is that I am always required to check if my transformation is bijective, and usually, I have no clue whether it is or not, I just state that yeah, it totally is .... but why? How do I check these things?
The function I am using in a current example is $$h(x,y) = (\sqrt{y}x,y)$$
on $\mathbb{R} \times (0,\infty)$.
 A: Given function $h: \mathbb R\times(0, \infty) \to \mathbb R \times (0, \infty)$ determined by $h\big((x,y)\big) = (\sqrt{y}x , y)$ we want to see if it is bijective. 
Let say $A = B = \mathbb R\times(0, \infty)$.
$h$ will be bijective iif $h$ is injective and exhaustive.
Injective:
Is $h$ injective? In other words, $\forall a, a'\in A, h(a)=h(a') \implies a = a'$?
$$h(a) = h(a') \implies (\sqrt{a_2}a_1, a_2) = (\sqrt{a_2'}a_1', a_2')$$
(We have supposed that $\exists a_1,a_2$ such as $a = (a_1, a_2)$ and also $\exists a_1',a_2'$ such as $a' = (a_1', a_2')$)
The implication follows $\sqrt{a_2}a1 =\sqrt{a_2'}a_1'$ and $a_2 = a_2'$. So, by substitution of the second equation in the first one, we have $\sqrt{a_2}a_1 = \sqrt{a_2}a_1'$, so $a_1 = a_1'$.
With this, we have that $a_2 = a_2'$ and $a_1 = a_1'$, so $a = a'$. We have proved that $h$ is injective.
Exhaustive:
We want to see if $\forall b \in B \exists a \in A$ such as $h(a) = b$.
Again, $\exists b_1,b_2$ such as $b = (b_1,b_2)$ and $\exists a_1,a_2$ such as $a = (a_1, a_2)$.
It follows
$$h(a) = b \implies (b_1, b_2) = (\sqrt{a_2}a_1, a_2)$$
so $b_1 = \sqrt{a_2}a_1$ and $b_2 = a_2$.
We already have $a_2$:$\quad a_2 = b_2$.
$a_1$ is not hard to obtain:
$$a_1 = \frac{ b_1 }{ \sqrt{a_2} } = \frac{ b_1 }{ \sqrt{b_2}}$$
If we join $a = (a_1, a_2) = \left(\frac{b_1}{\sqrt{a_2}}, b_2\right)$, we have found an $a$ that verifies our thesis, so $h$ is also exhaustive.
Bijective:
As $h$ is injective and exhaustive, we can finally end up that $h$ is bijective. QED.
