In the following graph, each node represents a computational problem. An arrow like A -> F indicates that there is a polynomial time Karp reduction from A to F. Observe that there could be more reductions than the ones indicated.
Suppose further that problem C is the (decision version of the) Independent Set problem.
According to the solution to the problem all problems belong to PSPACE, I don't understand why F belongs to this class. Why does this happen?
This just boils down to connectivity. If there is a path from the vertex $A$ to the vertex $B$, then the problem $A$ reduces to the problem $B$. There can be additional vertices on the path. Say, $A\rightarrow C\rightarrow B$, then you can reduce the problem $A$ to the problem $C$ first, then to the problem $B$.
In case of your particular graph, there is a path from $F$ to $C$: $F\rightarrow E\rightarrow B\rightarrow C$. This implies $F$ is reducible to the decision version of the Independent Set problem.
