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.