# Why does problem F belong to PSPACE?

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.