Let $V$ vector spaces. Prove $Hom(\mathbb{K},V)$ is isomorphic to $V$
I find a hint for this exercise:
Consider $\varphi\colon\operatorname{Hom}(\mathbb{K},V)\to V$ defined by linear form $f\colon\mathbb{K}\to V$, by $\varphi (f)=f(1)$
We need to show $\varphi$ is biyective.
For injectivity:
Suppose $\varphi (f)=\varphi (g)$ then $f(1)=g(1)$ this implies $f=g$.
In consequence $\varphi$ is injective.
For surjectivity:
Let $u\in V$ and $f\in Hom(\mathbb{K},V)$ then $\varphi(f)=f(1)=f(1.u)=1f(u)=f(u)=u$
I dont know it is the correct form of prove this. Can someone help me? (Is a little confused to me the injectivity and surjectivity of functions in dual spaces.)
What you wrote for surjectivity doesn't make a lot of sense. For instance, $f(u)$ is meaningless, since $f$ is supposed to evaluate on scalars.
Given $u\in V$, you define $f\in \text{Hom}\,(\mathbb K,V)$ by $f(\lambda)=\lambda u$. Then $$ \varphi(f)=f(1)=u. $$ So $\varphi$ is surjective.