Let $f:E\to F$ be linear, then prove that if $E=\ker f\oplus H$, then $H$ is isomorphic to $\operatorname{Im} f$ Let $f:E\to F$ be linear where $E,F$ are vector spaces and $\dim E<\infty$. Let $H$ be such that $E=\ker f\oplus H$, then I wish to prove that $H $ is isomorphic to $ \operatorname{Im} f.$
Here is what I have done:
If $f:E\to F$ is linear and $\dim E<\infty$, then $$\dim E=\dim \ker f+\dim \operatorname{Im} f. \;\;\;\;\;(1)$$ Since $E=\ker f\oplus H$, then by formula of four-dimension $$\dim E=\dim \ker f+\dim H.  \;\;\;\;\;(2)$$
From $(1)$ and $(2)$, we have that
$$\dim H=\dim \operatorname{Im} f.$$ 
Then, $H $ is isomorphic to $ \operatorname{Im} f.$ Please, I'm I right or wrong? If no, I would love to see better solutions.
 A: Yes, your proof is right. It can be seen as a direct consequence of the rank–nullity theorem, the Grassmann formula and the fact that 

Two finite dimensional vector spaces over a field $K$ are isomprphic iff they have the same dimension 

The proof of this is straightforward because if they are isomorphic there is an isomorphism $\phi$ that maps the base of one onto the base to the other so they have the same dimension. On the other hand if they have the same dimension they are both isomorphic to $K$ and so isomorphic to each other
A: Yes, this looks correct.
Since you have an explicit assumption that $E$ is finite-dimensional, this is probably also what you're supposed to do.
But the conclusion is also true in the infinite-dimensional case -- in fact you should be able to prove that the restriction of $f$ to $H$ is the isomorphism you're looking for, without counting dimensions at all.
A: The result is true regardless the dimension of $E$; that is, even when $E$ is infinite-dimensional:
$H$ is a complement of $\text{ker} f$, and all its complements are isomorphic to $V/\text{ker} f$. From this and applying the First Isomorphism Theorem, we get that
$$H \simeq V/\text{ker} f \simeq \text{Im} f.$$
