A bijective function from a $\mathbb{C}$-linear vector space to an $\mathbb{R}$-linear vector space

Question

Let $T:\mathbb{C} \to \mathbb{R}$ be the map defined by $T(z)=z+\overline{z}\,$.

For a $\mathbb{C}$-vector space $V$, consider the map $$\phi: \{f:V \to \mathbb{C} \,|\,f \text { is } \mathbb{C}\text{-linear} \}\to \{g:V \to \mathbb{R} \,|\,g \text { is } \mathbb{R}\text{-linear} \}$$ defined by $\phi(f)=T\circ f$. Prove that the map is bijective.

Assuming $\phi(f) = \phi(g)$ I get that the real parts of $f$ and $g$ are equal. How should I proceed to prove injectivity and surjectivity? Please help.

Answer 1

This is false. Let $V=\mathbb C$, $f(z)=z$ and $g(z)=\overline {z}$. Then $T\circ f=T\circ g$ but $f \neq g$

Answer 2

The injectivity of $\phi$ is not difficult (since $f\in \ker(\phi)$ if and only if $\operatorname{im}(f)\subseteq \text{i}\,\mathbb{R}$). The surjectivity of $\phi$ can be proven by defining the inverse $\psi$ of $\phi$ as follows. Let $$\big(\psi(g)\big)(v):=\dfrac{1}{2}\big(g(v)-\text{i}\,g(\text{i}v)\big)$$ for each $\mathbb{R}$-linear map $g:V\to \mathbb{R}$ and $v\in V$. Prove that $\psi$ is indeed the inverse map of $\phi$.