I need to prove that if a set of vectors $S$ is linearly independent, then $f(S)$ is also independent. Assuming f is the next homomorphism of vector spaces:
$$ f : V \rightarrow W$$
Demonstration attempt
We assume S is the next set of vectors:
$$ S = \left \{ u_{1}, u_{2},...,u_{n} \right \}$$
By definition of a linearly independent set, we know that the only solution to the equation is the trivial one:
$$\alpha_{1} u_{1} + \alpha_{2} u_{2} + ... + \alpha_{n} u_{n} = 0$$
Then,
$$f(\alpha_{1} u_{1}) + f(\alpha_{2} u_{2}) + ... + f(\alpha_{n} u_{n}) = 0 \Rightarrow \\ \alpha_{1} f(u_{1}) + \alpha f(u_{2}) + ... + \alpha f(u_{n}) = 0 \Rightarrow \\ \alpha_{1} = ... = \alpha_{n} = 0$$
It is a linearly independent set. $\blacksquare$
The result is false unless you have extra hypothesis on $f.$ The linear function $f(x) = 0$ already shows this.