I have the following problem:
$P$ is NP-hard, if and only if, there exists a NP-hard problem $Q$ such that $Q \preceq P$, i.e, $P$ is reducible to $Q$.
My attempt:
Let's show that $P$ is NP-hard. By hypothesis, we know that there exists a NP-hard problem $Q$ such that $Q \preceq P$, but $3SAT \preceq Q$. Hence, by transitivity of the relation $\preceq$, we have $3SAT \preceq P$. Therefore, $P$ is an NP-hard problem.
Now, let's prove the second implication. We know by definition of NP-hardness, that there exists an algorithm $M$ such that for all instances $\alpha \in$ 3SAT, we have $output_{M}(\alpha) \in P$. Now, consider the set $Q = \{ output_{M}(\alpha) \mid \alpha \in 3SAT \}$. Let $K$ be an algorithm such that maps every instance $b \in Q$ to $b$. Therefore, for all $\beta \in Q$, we have $output_{M}(\beta) \in P$. Hence $Q \preceq P$.
I feel that the second implication is not right. I would appreciate comments about my prove. Thanks! :)