I'm seeking clarification on the significance of commutative diagrams in understanding the analytic topology of smooth varieties.
In Aleksander Horawa's notes, a commutative diagram is used to demonstrate that a morphism between varieties is a continuous map on the analytic topologies (Remarks $1.3$):
$$ \begin{array}{ccc} X & \overset{\text{immersion}}{\longrightarrow} & \mathbb{A}^m \\ f\big\downarrow & \cdot & \big\downarrow{g} \\ Y & \underset{\text{immersion}}{\longrightarrow} & \mathbb{A}^n \end{array} $$
However, this seems like a straightforward statement to me. We know that a morphism is a polynomial from $ \mathbb{C}^m $ to $ \mathbb{C}^n $, which has been restricted to the variety $ X $. Hence, $ f^{-1}(U \cap Y) = f^{-1}(U) \cap f^{-1}(Y) = f^{-1}(U) \cap X $, where $ U $ is an open set with respect to the Euclidean topology on $ \mathbb{C}^m $.
I'm struggling to see the significance of these statements when it comes to the analytic topology, as open sets are simply given by $ X \cap O$, where $O$ is an open set in the Euclidean topology.
Another example is Proposition $1.1$, especially ($3$). These statements appear obvious to me, and it doesn't seem like they require such elaborate proofs.
I'm very interested to know if there's something I'm missing.