Dual Mapping Preserves Linear Independence if and only if Original Mapping is Surjective

Question

Here is my question:

Let $V$ and $W$ be finite-dimensional vectors spaces over a field $F$ and $f:V \rightarrow W$ a linear map.

Show that $f$ is surjective if and only if the image under $f^*:W^* \rightarrow V^*$, the dual map, of every linearly independent set of $W^*$ is linearly independent in $V^*$.

I would like to say that a map is surjective if and only if its dual is injective - is this true (and how to prove if so!)?

In the first part of the question, I had to show (the fairly standard) {$f$ injective} $\iff$ {$f$ preserves linear independence}. So I would then be done in both directions since $f^*$ is also linear.

Answer 1

You have the right approach.

Suppose first $f$ is surjective. If $g\in W^*$ and $f^*(g)=0$, then $g(f(v))=0$ for all $v\in V$ and since $f$ is surjective this means $g(w)=0$ for all $w\in W$, hence $g=0$. Thus $f^*$ is injective.

Now suppose $f$ is not surjective, so its image is a proper linear subspace. Let $w\notin f(V)$ and extend $w$ and a basis for $f(V)$ to a basis $w=w_1,w_2,\ldots,w_n$ for $W$. Define $g\in W^*$ as projection onto $\operatorname{span}\{w\}$ w.r.t. this basis, i.e. $$g(\lambda w+\lambda_2w_2+\cdots+\lambda_n w_n)=\lambda$$ Then $g\ne 0$ but $g(f(v))=0$ for all $v\in V$ so $f^*(g)=0$, hence $f^*$ is not injective.

Answer 2

I actually determined the answer shortly after posting! Sometimes just typing up the question helps as it makes you really think about it! :)

Claim: {$f:V \rightarrow W$ surjective} $\iff$ {$f^*:W^* \rightarrow V^*$ injective}, where $V$ and $W$ are as in the question.

Proof of Claim: "$\Rightarrow$" Suppose that $f$ is surjective. For $w \in W^*$, {$w \circ f = 0 \ \Rightarrow \ w=0$} (Indeed: Suppose not, say $w(x) = y \neq 0$; $f$ is surjective, so choose $z$ such that $f(z) = x$; then $(w \circ f)(z) = w(f(z)) = w(x) = y \neq 0$, which is a contradiction.] Thus $\ker(f^*) = \{0\}$, so $f^*$ is injective (as $f^*(\alpha) = \alpha \circ f$, by definition).

"$\Leftarrow$" Suppose that $f^*$ is injective. Write $f = i \circ s$ with $i$ injective and $s$ surjective, and both linear. Since $f^* = s^* \circ i^*$ is injective (by assumption), it follows that $i^*$ is injective. Since $V$ and $W$ are finite-dimensional, and the dimension of the dual space is the same as of the original space, $i$ must be surjective, since linear also. Thus $f$ is also surjective. $\Box$

Thus $\{ f \ \text {is surjective} \} \iff \{ f^* \ \text {is injective} \} \iff$ $\{$the image under $f^*:W^* \rightarrow V^*$, the dual map, of every linearly independent set of $W^*$ is linearly independent in $V^*$$\}$, as required. $\Box$

(I believe that this is correct!)