Time complexity to check if there is an edge between two nodes in an adjacency list

Question

I know that the time required to check if there exists an edge between two nodes in an adjacency matrix is $O(1)$ because we can access it directly (i.e: $M[i][j]$). However, I didn't really get why time complexity in an adjacency list to check would be $O(|V|)$, where $V$ represents the vertices. Shouldn't it be $O(|E|)$ instead? Because in the worst case, it needs to traverse through all of the elements (edges) in a linked list of a certain vertex.

Answer 1

In an adjacency list each vertex $u \in V$ is associated with a list of adjacent vertices.
Given a graph $G=(V,E)$, in order to check if the edge $(u,v) \in E$ you need to check whether $v \in \text{adjacent[u]}$.
A node can have at most $O(|V|)$ neighbors, from here the complexity follows.

Answer 2

Every vertex at most can have an edge with every other vertex which is $O(|V|)$ elements in the adjacency list per vertex. $|V|$ is just a tighter bound which has to be true due the first claim.