# Why these two field extension have the same index?

Let $h:C\rightarrow C'$ be nonconstant morphism over $k$ between two algebraic curves over $k$, and let $h^*:\bar{k}(C')\rightarrow \bar{k}(C)$ be the pullback of $h$ given by $f\mapsto f\circ h$.

My question is, why the following holds:

$$[k(C):h^*(k(C'))]=[\bar{k}(C):h^*(\bar{k}(C'))]$$

?

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