If a map between positive operators $\Phi: X \rightarrow Y$ is also completely positive, it is true that $\Phi\otimes I_A$ is also a positive map for any choice of ancilla operator space $A$.
That is, for any positive semidefinite operator $\rho \in X$, my map gives me a positive semidefinite operator $\Phi(\rho) = \rho' \in Y$. But my state could be entangled i.e. I could have had $\rho = Tr_A(\sigma)$ for some positive semidefinite $\sigma \in X\otimes A$. In this case, the map still works $(\Phi\otimes I_A) \sigma_{XA} = \sigma'_{XA}$ and gives a positive semidefinite output.
My question is regarding the dimension of $A$. I have seen proofs only consider the dimension of $A$ to be equal to the dimension of $X$. Why is it not required to consider a larger ancilla?
EDIT: To see the proof of what is written in the accepted answer, see https://en.wikipedia.org/wiki/Choi%27s_theorem_on_completely_positive_maps