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Why is the transpose map defined on dual space as follows is called the pullback map ?

$$T^t:V' \to V'$$ defined by $$T^t(f)=f\circ T $$

Moreover, why is it called the Pullback of $f$ by $T$. Any intuitive argument will be useful to me rather than technical details as I don't have much knowledge on linear algebra.

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More generally, suppose we have a linear transformation $T : V \to W$. Now let $f \in W^*$, the dual of $W$. That is, we have

$$V \xrightarrow{T} W \xrightarrow{f} \mathbb{F}$$

where $\mathbb{F}$ is the base field of $V$ and $W$. You can 'pull back' $f$ to be defined on $V$, rather than $W$, but still mapping into $\mathbb{F}$. This is acheived by precomposing with $T$ which gives an element of $V^*$. That is why we call the map $W^* \to V^*$, $f \mapsto f\circ T$ the pullback.

Heuristically, because we have a linear transformation $T$ which maps from $V$ to $W$, you can think of $V$ as the starting point, and $W$ as the destination. If we then have some object defined on $W$ (like an element of the dual space), we can try and pull it back from the destination to the starting point by somehow using $T$.

If I knew how to draw commutative diagrams, I'd include one here.

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