I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
3.94 Definition dual space, $V'$
The dual space of $V$, denoted $V'$, is the vector space of all linear functionals on $V$. In other words, $V'=\mathcal{L}(V,\mathbb{F})$.
Exercise 3.F.5
Suppose $V_1,\dots,V_m$ are vector spaces. Prove that $(V_1\times\dots\times V_m)'$ and $V_{1}^{'}\times\dots\times V_{m}^{'}$ are isomorphic vector spaces.
If $V_1,\dots,V_m$ are finite-dimensional vector spaces, then I can prove that $(V_1\times\dots\times V_m)'$ and $V_{1}^{'}\times\dots\times V_{m}^{'}$ are isomorphic vector spaces as follows:
My proof:
By 3.95 on p.101, $V_{1}^{'},\dots,V_{m}^{'}$ are finite-dimensional vector spaces and $\dim V_1=\dim V_{1}^{'},\dots,\dim V_m=\dim V_{m}^{'}$ hold.
By 3.76 on p.92, $V_1\times\dots\times V_m$ and $V_{1}^{'}\times\dots\times V_{m}^{'}$ are finite-dimensional vector spaces and $\dim(V_1\times\dots\times V_m)=\dim V_1+\dots+\dim V_m$ and $\dim(V_{1}^{'}\times\dots\times V_{m}^{'})=\dim V_{1}^{'}+\dots+\dim V_{m}^{'}$ hold.
By 3.95 on p.101, $(V_1\times\dots\times V_m)^{'}$ is a finite-dimensional vector space and $\dim (V_1\times\dots\times V_m)=\dim (V_1\times\dots\times V_m)^{'}$ holds.
So, $\dim (V_1\times\dots\times V_m)^{'}=\dim (V_1\times\dots\times V_m)=\dim V_1+\dots+\dim V_m=\dim V_{1}^{'}+\dots+\dim V_{m}^{'}=\dim(V_{1}^{'}\times\dots\times V_{m}^{'})$.
By 3.59 on p.82, $(V_1\times\dots\times V_m)'$ and $V_{1}^{'}\times\dots\times V_{m}^{'}$ are isomorphic vector spaces.
Best Answer
For $i\in \{1,\ldots,m\}$, let $p_i:V_1\times \ldots \times V_m \to V_i$ be the canonical linear surjections and $\iota_i:V_i\to V_1 \times \ldots \times V_m$ the canonical linear injections. We know that $p_i\circ \iota_j = 0$ if $i\neq j$ and $\rm{id}_{V_i}$ if $i=j$ and that : $$\rm{id}_{V_1\times\ldots\times V_m} = \sum_{i=1}^m \iota_i\circ p_i$$
Define a map $\Phi : V_1'\times \ldots \times V_m' \to (V_1\times \ldots \times V_m)'$ by the formula :
$$\forall f_1\in V_1', \ldots , f_m\in V_m',\Phi(f_1,\ldots,f_m) =\sum_{i=1}^m f_i\circ p_i$$
This does define a linear map, let us show that it is an isomorphism.
Let $f_1,\ldots,f_m$ such that $\Phi(f_1,\ldots,f_m) = 0$. Then, for $i\in\{1,\ldots,m\}$, we have : \begin{align} 0 &= \Phi(f_1,\ldots,f_m)\circ \iota_i \\ &= \sum_{j=1}^m f_j\circ p_j\circ \iota_i \\ &= f_i \end{align} Therefore $(f_1,\ldots,f_m) = 0$ and we see that $\Phi$ is injective.
Let $g\in (V_1\times \ldots \times V_m)'$ . For $i\in \{1,\ldots,m\}$ let $f_i = g\circ\iota_i \in V_i'$.
Then, we can compute : \begin{align} g &= g\circ\left(\sum_{i=1}^m \iota_i\circ p_i\right) \\ &= \sum_{i=1}^m g\circ \iota_i\circ p_i \\ &= \sum_{i=1}^m f_i\circ p_i \\ &= \Phi(f_1,\ldots,f_m) \end{align}
Therefore $\Phi$ is surjective.