# Reduction with NPH

I have a question in complexities that I could not do.

There will be D, E, F, three languages belonging to NPH. Suppose that the reductions exist $$D \leq _P E$$ and $$E \leq _P F$$. Determine which of the following statements is correct:

1. Inevitably there is a reduction $$F \leq _P D$$
2. If there was an $$F \leq _P D$$ reduction then F, E and D are all in the NPC class.
3. Even if there was an $$F \leq _P D$$ reduction, it is still possible that $$D \notin NP$$.
4. if $$F \in CoNP$$ then $$NP \neq CoNP$$
5. None of the above claims are true.

I think the answer is probably 2, because there are reductions between all the problems. So everyone should be in $$NPH$$, and if a problem is in $$NPH$$ then it is also in $$NPC$$. I can not understand why 4 is not true, it also seems logical.

if a problem is in $$NPH$$ then it is also in $$NPC$$
This statement is incorrect. A language is in $$NPC$$ if it is in $$NPH$$ and it is in $$NP$$.
Answer 4 is incorrect as well since that would imply that any language $$L\in NP$$ can be reduced to $$F$$, hence $$L\le_p F$$ and since $$F\in coNP$$ then $$L\in coNP$$. Hence, $$NP\subseteq coNP$$. Its not hard to show the other way around, and conclude that this means $$NP=coNP$$.
Answer 3 is correct. There are $$NPH$$ problems that is not in $$NP$$ (for example, take any $$NEXP$$-complete problem). If you choose $$D=E=F\in NPH\setminus NP$$ then the reductions are trivial, but they are still not in $$NP$$.