How Reduction works in proving NP-Hard?

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 found a 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.