# 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. Oct 2, 2013 at 23:29
• You can still extend a basis of $L(V)$ to a basis of $W$. Oct 2, 2013 at 23:30
• I do not understand the words that "that every injection is a section and similarly with surjections"…… Oct 2, 2013 at 23:34
• @Knight See my answer. Oct 3, 2013 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? Oct 3, 2013 at 11:57
• well without using a basis no Oct 3, 2013 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. Oct 3, 2013 at 12:58
• Can somebody please explain me why we are allowed to write $W \cong V \oplus U$ for some $U$? Thanks! Mar 22, 2014 at 10:53
• If $f : V \to W$ is injective and linear, $V$ is isomorphic to a subspace of $W$. But every subspace of a vector space is a direct summand; this is because we can extend bases from subspaces to the whole space: the 'rest' of the basis generates the other summand. Jul 21, 2021 at 23:12