Importance of the homogeneity assumption in definition of linear map Let $V$ and $W$ be vector spaces over field $F$.  A function $f: V \rightarrow W$ is said to be linear if for any two vectors $x$ and $y$ in $V$ and any scalar $\alpha\in F$, the following two conditions are satisfied:


*

*$f(x + y) = f(x) + f(y)$

*$f(\alpha x) = \alpha f(x)$


Let $F$ be a field of real numbers. Is it possible to construct $f$ such that the first condition is satified but not the second one?
 A: Yes, it is possible to construct examples of $f$ that are additive but not homogeneous. However, the construction won't be explicit and depends on the axiom of choice. See here for details.
If you want an explicit example of an additive map that is not homogeneous over a different field, let $\mathbb{F} = \mathbb{C}$ and consider the map $\varphi \colon \mathbb{C} \rightarrow \mathbb{C}$ given by complex conjugation:
$$ \varphi(z) = \varphi(x + iy) = \overline{z} = x - iy. $$
The map $\varphi$ is additive and $\mathbb{R}$-homogeneous but not $\mathbb{C}$-homogeneous as $\varphi(az) = \overline{az} = \overline{a}\overline{z} = \overline{a}\varphi(z)$.
A: A function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ is known as a Cauchy additive function. This would be a special case example when $V=W=\mathbb{R}$. Assuming the axiom of choice, there do exist such functions without the property that $f(\alpha x)=\alpha f(x)$, but such functions are incredibly ugly, for example they aren't continuous anywhere.
The proof that such functions exist is non-constructive (because it uses the axiom of choice). This means that you can't actually construct an example of one of them. 
A: Cannot be omitted. A homomorphism of vector spaces must preserve four operations: addition of vectors, multiplication with scalar, fixing zero vector, taking the inverse of vector.
It is necessary and sufficient that a linear mapping satisfy said axioms.
