Is the finite dimension required in this proof? Let $V$ and $W$ be vector spaces over a field $K$. If a linear map $L:V \rightarrow W$ is surjective then its dual is injective.
If $V$ and $W$ are finite dimensional then the converse holds, i.e.  $L^*:W^* \rightarrow V^*$ injective implies $L$ surjective.
I have proved both statements but I don't see where I used the finite dimensional requirement for the second. Here is my proof:
Assume $L$ is not surjective, say the element $e_i$ of the basis of $W$ is not in the image of $L$.  Take its corresponding dual $\alpha_i \in W^*$, then $L^*(\alpha_i)=\alpha_i \circ L =0$ so the kernel of $L^*$ is not 0 and therefore $L^*$ is not injective.
 A: My first argument has to be changed as the comments of @Julian Rosen show. Finite dimension is not needed:
Take a basis $B'$ of the image of $L$ in $W$ and complete it to a basis $B$ of $W$ (by assumption $B \setminus B'$ is not empty). Define the linear functional $\alpha$ by $\alpha(e)=1$ where $e\in B\setminus B'$ and $\alpha(v)=0$ for $v \in B'$ (and extend linearly). 
Then $L^*(\alpha)=\alpha \circ L = 0$, and $\alpha$ is not the $0$ linear functional, hence $L^*$ is not injective.
For the proof of $L^*$ surjective iff $L$ injective see the last answer of:
Are injectivity and surjectivity dual?
A: I think Pete L. Clark's answer is intuitive and I don't mean to obfuscate the problem, but could I add some abstract nonsense (which obviously one could ignore)?
In the category of vector spaces, we can easily show every mono (injective) and epi (surjective) splits. The dual construction is faithfully functorial (another straightforward exercise), hence it both preserves and reflects split monos and split epis, i.e. monos iff the dual is epi and epi iff dual is mono.
Thank you for the indulgence.
