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Let $F$ be a field and $V,W$ finite-dimensional vector spaces over $F$.

Let $f:V\rightarrow W$ a $F$-linear mapping.

We have to show that $f$ is surjective if and only if for each generator set $S$ of $V$ the Image $f(S)$ is a generator set of $W$.

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When $f$ is surjective it holds that $\dim im(f) = \dim W$ and also $im (f)=W$, or not?

We have that $\forall v\in V$ : $v=\sum_{i=1}^na_is_i, \ s_i\in S$.

Then $f(v)=\sum_{i=1}^na_if(s_i)$, since $f$ is linear.

How can we continue? How do we use the fact that $f$ is surjective?

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Hint: Prove that for all subset $S \subset V$ we have

$$\langle f(S)\rangle = f(\langle S \rangle).$$

If $f$ is surjective and $S$ is a generating set of $V$, then the RHS is $f(V) =W$, hence $f(S)$ is generating.

The converse is trivial upon taking for $S = V$, then $f(S) = f(V)$ is a generating subspace hence $f(V) =W$.

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    Could you explain to me further how we conclude that $f(S)$ is generating? $$$$ If $f$ is surjective und $y \in W$, there is a $x \in V$ with$f(x) = y$. Can we write $x$ as a linear combination of teh elements of $M$ : $x=\sum_i a_im_i$, where $m_i\in M$ ? – 2017-01-19
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    Ok, let $y \in W$. And let $x \in V$ be such that $f(x) = y$. Then write $x = \sum_{i}{a_i s_i}$. Now apply $f$ to get $y = f(x) = \sum_{i}{a_if(s_i)}$. So you have written $y$ as a linear combination of elements in $f(S)$. – 2017-01-19
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    Ah ok!! Thank you very much!! :-) – 2017-01-20