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

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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