Proof verification: almost linear map is injective iff the kernel only contains zero. $f:V\to W$ is a $\mathbb{R}$-$\mathbb{C}$-linear map between a $\mathbb{R}$-vector space $V$ and a $\mathbb{C}$-vector space $W$ if
$$
f(av+bw)=af(v)+bf(w):=(a+i0)f(v)+(b+i0)f(w)
$$
for any $a,b\in\mathbb{R}$ and $v,w\in V$.

Theorem: A $\mathbb{R}$-$\mathbb{C}$-linear map is injective iff the kernel only contains zero.

Proof:
 Assume that $f$ is injective and note that
    \begin{align*}
 f(\mathbf{0})= f(0\cdot \mathbf{0}) = 0f(\mathbf{0}):=(0+i0)f(\mathbf{0}) = \mathbf{0},
 \end{align*}
    proving that $\mathbf{0}\in \mathrm{ker}(f):=\{v\in V: f(v)=\mathbf{0}\}$, and by injectivity of $f$ we have that $\mathrm{ker}(f)$ must be a singleton, hence $\mathrm{ker}(f)=\{\mathbf{0}\}$. Conversely assume that $\mathrm{ker}(f)=\{\mathbf{0}\}$ and consider any two elements $v_1,v_2\in V$ such that $f(v_1)=f(v_2)$. Then $f(v_1-v_2)=f(v_1)-f(v_2)=\mathbf{0}$, implying that $v_1-v_2\in \mathrm{ker}(f)$, but this must entail that $v_1-v_2=\mathbf{0}\iff v_1=v_2$, since $\mathrm{ker}(f)=\{\mathbf{0}\}$. 
QED.
This proof is a straight forward replication of the true linear case, but I can't find any problems with it in this case. Am i correct that I can extend the well known theorem, to what i defined as $\mathbb{R}$-$\mathbb{C}$-linear maps above?

Your proof is fine. 
Here is another reason it should work out the same: The vectors in a complex vector space also form a real vector space under vector addition and scalar multiplication by reals. Your "almost linear" maps are then linear maps to this corresponding real vector space. Since you already have this result for linear maps between real vector spaces, it still holds here.
