Isomorphism of groups by commuting the kernel I'm looking to show whether the following is a homomorphism and by commuting the kernel if it is an isomorphism:
$G = H = ((0,\infty),×), f(x) = x^2$
Firstly do I check that $f(x+y)=f(x)f(y)$ or check $f(xy)=f(x)f(y)$
I checked the following...
So under multiplication I have that 
$f(xy)=(xy)^2=x^2y^2=f(x)f(y)$ hence we have a surjective homomorphism:
I know the following definition for a kernel;
The kernel of a group homomorphism $\phi: G \rightarrow H$ is the subset $ker\phi := ({g \in G | \phi(g) = e_H})$. 
I don't really understand how to apply my example to this to see if the above homomorphism is an isomorphism. 
I presume the kernel should show injectivity hence leading to a bijection and thus an isomorphism?
My assumption would be that the kernel would be 
ker $\phi :=(x \in G | \phi(-1)=1)$
Edit
if $G$ and $H$ are groups, an isomorphism $f : G → H$ is a bijection $f : G → H$ such that, for all $g_1,g_2 \in G, f(g_1g_2) = f(g_1)f(g_2)$. 
 A: I assume that $f:G\to H$? If that's the case, then the group operation is $\times$. And so you have to check that $f(x\times y)=f(x)\times f(y)$. And thus you correctly checked that $f$ is a group homomorphism. The fundamental reason is that the standard real numbers multiplication is commutative, and so $(xy)^2=xyxy=xxyy=x^2y^2$.
$f$ is also surjective because for any $y\in (0,\infty)$ the equation $f(x)=y$ has a solution, namely $x=\sqrt{y}$.
When we talk about kernels, we have to know neutral elements on both sides. So the neutral element of $G=H$ is $1$. And so we can replace $e_H$ (which stands for abstract neutral element of $H$) by $1$:
$$\ker f=\{x\in G\ |\ f(x)=1\}=f^{-1}(\{1\})$$
And it is well known that a homomorphism is injective if and only if its kernel is trivial (meaning it contains only the neutral element).
So what exactly is the kernel of $f$? We need to solve the equation $f(x)=1$, i.e. $x^2=1$. And it is well known that this has only one solution in $(0,\infty)$, namely $x=1$. Therefore $\ker f=\{1\}$ is trivial. And so $f$ is injective.
