Real forms of a complex vector space correspond exactly to conjugations

For a conjugation $\sigma$, the subspace $W^{\sigma }$ is a real form of $W$: by [L1] the complexification of $W^{\sigma }$ recovers $W$ through the multiplication map, which is exactly the defining condition.

The fixed points of ${\sigma }_{\mathrm{can}}$ on $V_{\mathbb{C}}$ are the embedded copy of $V$: by [L4] an element is uniquely $1\otimes A+i\otimes B$, and ${\sigma }_{\mathrm{can}}(1\otimes A+i\otimes B)=1\otimes A-i\otimes B$ by [L2], which equals itself exactly when $i\otimes B=-i\otimes B$, hence $B=0$.

For a real form $V$ with isomorphism $\theta :V_{\mathbb{C}}\to W$, define ${\sigma }_V(w)=\theta ({\sigma }_{\mathrm{can}}({\theta }^{-1}w))$. It is additive and conjugate-linear because $\theta$ is complex-linear and ${\sigma }_{\mathrm{can}}$ has these properties by [L2], and it is an involution because ${\sigma }_{\mathrm{can}}^{}=\mathrm{id}$.

If $V=W^{\sigma }$ came from a conjugation $\sigma$, then ${\sigma }_V=\sigma$: for $w=A+iB$ with $A,B\in W^{\sigma }$, step 1.3 gives ${\sigma }_V(w)=\theta (1\otimes A-i\otimes B)=A-iB$, while $\sigma (A+iB)=A+\bar{i}B=A-iB$ by [L2].

If ${\sigma }_V$ came from a real form $V$, then $W^{{\sigma }_V}=\theta (\{1\otimes A:A\in V\})=V$: the fixed points of ${\sigma }_V$ in $W$ are the $\theta$-images of the fixed points of ${\sigma }_{\mathrm{can}}$, which step 1.2 identifies with the embedded copy of $V$, and $\theta (1\otimes A)=A$ by the real-form condition.

Steps 1.1, 2.1 and 2.2 show that the two assignments compose to the identity in both orders, so they are inverse bijections.