Let \(V\) be a vector space over a field \(F\). The evaluation map \(e\colon V \to (V^\vee)^\vee\) is defined by \(e(v)(f) \da f(v)\) for \(v\in V\) and \(f\in V^\vee\).
Prove that \(e\) is an injection.
Prove that \(e\) is an isomorphism if and only if \(V\) is finite dimensional.
Solution
Write \(V^\vee=\operatorname{Hom}_F(V,F)\) for the algebraic dual. We use basis existence and basis extension, valid for arbitrary vector spaces by Zorn’s lemma [@DF04].
1. The map \(e:V\to V^{\vee\vee}\) is linear and injective.
For fixed \(v\), evaluation \(f\mapsto f(v)\) is linear on \(V^\vee\). For \(a,b\in F\) and \(v,w\in V\), \[ e(av+bw)(f)=f(av+bw)=a e(v)(f)+b e(w)(f) \] for every \(f\), so \(e\) is linear. If \(v\ne0\), extend \(\{v\}\) to a basis of \(V\). Assign value one to \(v\) and zero to the other basis vectors, and extend linearly to a functional \(f\). Then \(e(v)(f)=1\), so \(e(v)\ne0\). The kernel is therefore zero.
2. When \(V\) is finite-dimensional, \(e\) is onto.
Choose a basis \(v_1,\ldots,v_n\) and let \(\varepsilon_i\) be its coordinate functionals. Every \(f\in V^\vee\) satisfies \(f=\sum_{i=1}^n f(v_i)\varepsilon_i\) by evaluation on the basis. For \(\Phi\in V^{\vee\vee}\) set \(v=\sum_{i=1}^n\Phi(\varepsilon_i)v_i\). Then \[ \Phi(f)=\sum_{i=1}^n f(v_i)\Phi(\varepsilon_i)=f(v)=e(v)(f). \] Hence \(\Phi=e(v)\), proving surjectivity. If \(V=0\), the same statement uses the empty basis and all three spaces are zero. Together with step 1, this proves the forward implication in part (2).
3. When \(V\) is infinite-dimensional, \(e\) is not onto.
Choose a basis \((v_i)_{i\in I}\) with \(I\) infinite. Sending \(f\) to \((f(v_i))_{i\in I}\) identifies \(V^\vee\) with the full function space \(F^I\): every assignment on the basis extends uniquely by finite linear sums. Let \(D\subset F^I\) be the subspace of finite-support functions, and let \(\mathbf1\) be the constant-one function. Since \(I\) is infinite, \(\mathbf1\notin D\).
On \(D+F\mathbf1\) define \(\lambda_0(d+c\mathbf1)=c\). The sum is direct: a nonzero constant function cannot have finite support. Thus this rule is well defined and linear. Extend a basis of this subspace to a basis of \(F^I\) and assign value zero on the added basis vectors. This extends \(\lambda_0\) to a linear functional \(\lambda:F^I\to F\) with \(\lambda(D)=0\) and \(\lambda(\mathbf1)=1\).
Via the identification above, \(\lambda\) is an element of \(V^{\vee\vee}\). Suppose it equals \(e(v)\). Write \(v=\sum_{i\in J}a_i v_i\) for a finite set \(J\). The coordinate functional \(\varepsilon_i\) corresponds to the function supported at \(i\) with value one, which belongs to \(D\). Thus \(0=\lambda(\varepsilon_i)=\varepsilon_i(v)=a_i\) for every \(i\in J\). Hence \(v=0\), which would imply \(\lambda=0\), contrary to \(\lambda(\mathbf1)=1\). This constructs a double-dual element not in the image and proves the converse implication.