Exponential on Real Numbers is Group Isomorphism/Proof 2

Theorem
Let $\left({\R, +}\right)$ be the additive group of real numbers.

Let $\left({\R_{> 0}, \times}\right)$ be the multiplicative group of positive real numbers.

Let $\exp: \left({\R, +}\right) \to \left({\R_{> 0}, \times}\right)$ be the mapping:
 * $x \mapsto \exp \left({x}\right)$

where $\exp$ is the exponential function.

Then $\exp$ is a group isomorphism.

Proof
From Real Numbers under Addition form Abelian Group, $\left({\R, +}\right)$ is a group.

From Positive Real Numbers under Multiplication form Abelian Group, $\left({\R_{> 0}, \times}\right)$ is a group.

We have that for all $y \in R_{> 0}$ there exists $x = \ln \left({y}\right) \in R$ which satisfies $\exp \left({x}\right) = y$.

Thus $\exp$ is a surjection.

Then from Exponential is Strictly Increasing and Strictly Convex, $\exp$ is strictly increasing on $\R$.

From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $\exp$ is an injection.

Therefore, $\exp$ is a bijection.

Let $x, y \in R$.

From Exponent of Sum:
 * $\exp \left({x + y}\right) = \exp \left({x}\right) \exp \left({y}\right)$

So $\exp$ is a homomorphism and a bijection.

It follows by definition that $\exp: \left({\R, +}\right) \to \left({\R_{> 0}, \times}\right)$ is an isomorphism.