Naming of contravariant vector field and covariant vector field

I don’t know why contravariant and covariant vector field are named as such. contravariant literally means going against changing, or changing in the opposite way, covariant literally means changing with something in the same way.

For example, vector field and its dual are (in classical terminology) seems contravariant and covariant vector field of the first order. I don’t see how vector field and its dual bear the literal meanings of the two words. Wiki says that they have something to do with their relations to basis vector, but I still can’t figure it out.
Is there any reason or historical explanation for that naming?

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I see so

1. covariant vector is an element in dual space, it’s transformation of or action on vector, but (for first order tensor this transformation of vector is inner product so) it can also be regarded as a transposed vector, and visualized in the same space (but it’s not in the same space) as the upenn link does. The key is to recognize the covariant vector in the above $e^j e_i$ Is not ‘vector‘ $e^j$ but $e^j$ as a transformation or transposed vector (but we need to realize that only when it act on a vector it’s covariant vector).

2. there are two dual: Space of covariant vector, dual space. Its basis can be any proper set of covariant vectors; dual vector, the special basis of dual space $e^j$ that satisfy that its action on basis of vector space produces $delta_{i, j}$.

3. formally contravariant means inverse of basis transformation matrix: after changing basis by transformation matrix A, the same vector (=components * basis) equals (components transformed by inverse transformation $A^{-1}$ times $A$ transformed basis, i.e. the new basis $e’_i$ ), the result is easier to understand from perspective of matrix operation as insertion of $A^{-1}A$. For dual basis it’s similar. Transformation $B$ of dual basis that satisfies ‘duality’ requirement needs to be inverse $A^{-1}$ of A too,—that is $delta_{i, j}= e^j e_i = e’^j e’_i = (e^j B)(A e_i)$, where $e’^j , e’_i$ are transformed basis, so $B= A^{-1}$—so transformation of components of dual basis (the components can be easily understood if we view covariant vector as a transposed vector) seems to be just $A$, in this sense it’s covariant.