Prove that $(V_{1} \times .......\times V_{m})'$ and $V'_{1} \times .......\times V'_m$ are isomorphic vector spaces

Question

Suppose $V_1, \dots, V_m$ are vector spces. Prove that $(V_{1} \times \dots\times V_{m})'$ and $V'_{1} \times \dots\times V'_m$ are isomorphic vector spaces, where $V'_i$ represents the dual space of the vector space $V_i$.

There is a solution out on the net already (Question #5): http://linearalgebras.com/3f.html, but I am having trouble interpreting the definition of their isomorphism: $$\varphi(f)=(P’_1f,\cdots,P’_mf).$$

Where did this $f$ come from?

I ask because by the definition of the dual map: $$T': W'\to V' \\ \phi \stackrel{T'}\longrightarrow \phi\circ T$$

I would equate $P_i = T$ and $\varphi = T'$. So how is the isomorphism an element of $V'_{1} \times \dots\times V'_m$?

Answer 1

You want to define an isomorphism $\phi\colon (V_1\times \ldots\times V_m)'\to V_1'\times\ldots\times V_m'$, so for any given element of $(V_1\times \ldots\times V_m)'$, you have to define its image under $\phi$. This "any given element" is $f$. So $f$ is an arbitrary element of $(V_1\times \ldots\times V_m)'$, which means it is an arbitrary linear map $V_1\times \ldots\times V_m\to k$. For each $i$, we have the injection on the $i$th component $P_i\colon V_i\to V_1\times \ldots \times V_m$, and hence the duality $P_i'\colon (V_1\times \ldots \times V_m)'\to V_i'$. Thus from $f$ we can form $P_i'f\in V_i'$. Then the tuple $(P_1'f,\ldots, P_m'f)$ is obviously in $V_1'\times \ldots\times V_m'$.

(I am just confused by the naming $P_i$ for the canonical inclusions and not for the canonical projections.)

Answer 2

You don't want the isomorphism to be an element of $V_1'\times\cdots\times V'$. An isomorphism between $(V_1\times\cdots\times V)'$ and $V_1'\times\cdots\times V'$ is a bijective linear homomorphism between these two spaces. They're exhibiting such a map by defining how it acts on an arbitrary element $f$. So the answer to your question "Where did this $f$ come from?" is: out of nowhere. It's just an arbitrary element being mapped to provide a definition of $\phi$.