A problem $X$ is $NP$-Hard if for all $Y \in NP$, $Y \leq_P X$. Further, if a problem $Z$ is $NP$-Complete, and $Z \leq_P X$, then I can prove (rather mechanically) that $X$ is $NP$-Hard.
I also found that, for a given X, say TSP, to prove it is $NP$-Hard, we often find a ($NP$-Complete) Z say $HAM-CYCLE$ (Hamiltonian-Cycle), and try to show that $HAM-CYCLE$ can be reduced to a $TSP$ in polynomial time.
My confusion is, why don't we try to show that for each instance of $TSP$, there exists a corresponding instance of $HAM-CYCLE$. Specifically, what if there exists an instance of $TSP$, for which there is no corresponding instance in $HAM-CYCLE$! In this case, how can the prior knowledge about the hardness of $HAM-CYCLE$ help in inferring on TSP's hardness!
Note: I also had similar concerns with proving $NP$-complete class. However, since all $NPC$ are also $NP$, I felt, similar reduction of a known NPC problem to a given problem, say Q, works. However, for $NP$-hard case, a given problem X need not to be $NP$ at all.