Injectivity of the dual map Suppose V and W are vector spaces of  possibly finite and infinite dimension over a field K. Show that if a linear map $L : V → W$ is surjective the its dual is injective.
Also prove the converse of the last implication.
Well when V,W are finite spaces i can prove it and i understand that dimension is not necessary if i want prove surjective implies injective. Other way injective implies surjective when is finite i take a basis ${e_1,….,e_n}$ of V then ${Le_1,….,Le_n}$ is l.i. in W so we can extend a basis ${Le_1,….,Le_n,w_1,…,w_k}$ of W and define:$f: W → K$ by $f(Le_i) = g(w_i)$ and $f(w_i)=0$ then $L^*: V → K$ and$L^*(f)(e_i) = (fL)(e_i) = f(Le_i)=g(e_i)$ then $L^*(f)=g$
But what happen if V and W is infinite dimension?.
 A: *

*If $f : V \to W$ is surjective, then $f^* : W^* \to V^*$ is injective, since clearly $\omega \circ f = 0$ implies $\omega = 0$ for $\omega \in W^*$.

*If $f : V \to W$ is injective, then it has a section, i.e. a linear map $g : W \to V$ with $gf=1_V$. Then $f^* g^* = 1_{V^*}$, hence $f^*$ is surjective. Explicitly, we can write $W \cong V \oplus U$ and $f$ corresponds to the inclusion. Then $f^*$ corresponds to the projection $V^* \oplus U^* \to V^*$. This also shows: If $f^*$ is injective, then $U^* = 0$, hence $U=0$, and $f$ is an isomorphism.

*If $f^* : W^* \to V^*$ is injective, then write $f=ig$ with $i$ injective and $g$ surjective. Since $f^*=g^* i^*$ is injective, it follows that $i^*$ is injective. By 2. $i$ is an isomorphism. Hence $f$ is surjective.

*If $f^* : W^* \to V^*$ is surjective, then by 1. $f^{**} : V^{**} \to W^{**}$ is injective. Using the commutative diagram
$$\begin{array}{c} V & \rightarrow & W \\ \downarrow && \downarrow \\ V^{**} & \rightarrow & W^{**} \end{array}$$
and that $W \to W^{**}$ is injective, we see that $f$ is injective.
