Is there any decision problem not in NPH but can be reduced to a NPH problem?
Assuming you're using many-one reductions, $\emptyset$ and $\Sigma^*$ are not NP-hard (exercise: prove this!), but they can be reduced to any NP-hard problem. Therefore, the statement, "one can show that a problem&nbsp;$A$ is in NPH by giving a reduction from $A$ to a NPH problem&nbsp;$B$" is false.