# 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?.

-
Hint: use the fact the dual space functor is contravariant, and that every injection is a section and similarly with surjections. –  Alex Youcis Oct 2 '13 at 23:29
You can still extend a basis of $L(V)$ to a basis of $W$. –  Daniel Fischer Oct 2 '13 at 23:30
I do not understand the words that "that every injection is a section and similarly with surjections"…… –  Knight Oct 2 '13 at 23:34
@Knight See my answer. –  Alex Youcis Oct 3 '13 at 0:24

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

2. 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.

3. 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.

4. 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.

-
Out of curiosity, do you know how to prove "then $U^\ast = 0$, hence $U=0$" without using a basis? –  Daniel Fischer Oct 3 '13 at 11:57
well without using a basis no –  Knight Oct 3 '13 at 12:31
Or Zorn or something else equivalent to the axiom of choice. I hesitated to mention that lest somebody might think I had qualms about choice. –  Daniel Fischer Oct 3 '13 at 12:58
Can somebody please explain me why we are allowed to write $W \cong V \oplus U$ for some $U$? Thanks! –  Mathoman Mar 22 '14 at 10:53

Suppose first that $L:V\to W$ is surjective. Then, there exists some linear map $\sigma:W\to V$ such that $L\circ\sigma=\text{id}_W$. Then, note that if we denote the dual map of $T$ by $T^\vee$ we have that

$$\text{id}_{W^\vee}=(\text{id}_W)^\vee=(L\circ \sigma)^\vee=\sigma^\vee\circ L^\vee$$

From this you can see that $L^\vee$ has a left inverse, and thus $L^\vee$ is injective.

Conversely, suppose that $L^\vee$ is injective. Suppose that $L(v)\ne w$ for all $v\in V$. Extend $\{w\}$ to a basis $\{w\}\cup\{w_\alpha\}$. Define a linear functional $\varphi\in W^\vee$ by $\varphi(w)=1$ and $\varphi(w_\alpha)=0$ for any $\alpha$. Note that $\varphi\ne 0$, but for any $v\in V$ we have that

$$L^\vee(\varphi)(v)=\varphi(L(v))=0$$

since $L(v)\notin \text{span}(w)$. Thus, $L^\vee(\varphi)=0$, and thus $\varphi$ is a non-trivial element of $\ker L^\vee$, which contradicts $L^\vee$'s injecivity. Thus, $L$ must be surjective.

-
Yeap i wrote that I proved surjective implies injective, the problem is injective implies surjective when V and W are infinite dimension the other ones i proved it. –  Knight Oct 3 '13 at 0:47
@Knight I fixed it. –  Alex Youcis Oct 3 '13 at 7:25
The proof is not correct. $L(v) \notin \langle w \rangle$ doesn't imply that $L(v) \in \langle w_{\alpha} \rangle_{\alpha}$. You have to adjust the definition of the basis. –  Martin Brandenburg Oct 3 '13 at 9:55