Let $T:V\rightarrow W$ a linear transformation, $i_w:W\rightarrow W^{**}$ and $i_v:V\rightarrow V^{**}$ canonical morphism of biduality. Let $T:V\rightarrow W$ a linear transformation, $i_w:W\rightarrow W^{**}$ and $i_v:V\rightarrow V^{**}$ canonical morphism of biduality.
Prove $i_w\circ T=T^{**}\circ i_v$
I' m very very confused with this exercise.
Suppose i need to prove this:
$i_w\circ T\subset T^{**}\circ i_v$ and $T^{**}\circ i_v\subset i_w\circ T $ but  i don't have idea of how to prove this. Can somone give me a hint?

Let's recall the definitions: if $T : V \to W$, then $T^{*} : W^* \to V^*$ is a linear map defined as $T^{*}(f) = f \circ T$ for all $f \in W^*$.
Then, $T^{**} : V^{**} \to W^{**}$ is defined as $T^{**}(g) = g\circ T$ for all $g \in V^{**}$.
Now, to prove $i_w\circ T=T^{**}\circ i_v$ let's first establish the domain and codomain of both sides. Since $T : V \to W$ and $i_W : W \to W^{**}$, we have $i_w\circ T : V \to W^{**}$. Similarly, since $i_v : V \to V^{**}$ and $T^{**} : V^{**} \to W^{**}$, we have $T^{**}\circ i_v : V \to W^{**}$, so both sides represent a function with the same domain and codomain.
Let $x \in V$ be arbitrary. Let's show that $ (i_w\circ T)(x)= (T^{**}\circ i_v)(x)$. This is an equality of two functions $W^* \to \mathbb{F}$, where $\mathbb{F}$ is the underlying field of your vector spaces. Therefore take an arbitrary $f \in W^*$, and let's show $ (i_v\circ T)(x)(f)= (T^{**}\circ i_v)(x)(f)$ in $\mathbb{F}$.
\begin{align}(i_w\circ T)(x)(f) &= (i_w(T(x))(f)\\
&= f(T(x))\\
&= (f\circ T)(x)\\
&= (i_v(x))(f\circ T)\\
&= (i_v(x))(T^*(f))\\
&= (i_v(x) \circ T^*)(f)\\
&= (T^{**}(i_v(x))(f)\\
&= (T^{**}\circ i_v)(x)(f)\\
\end{align}
Hence, $i_w\circ T=T^{**}\circ i_v$.
