If both $A$ and $B$ are NP-complete languages, would $A \times B$ be necessarily NP-hard? Here, $A \times B = \{\langle u,v \rangle \mid u\in A, v\in B\}$.
So my colleague's answer was no, it's not necessarily NP-hard because you can verify in polynomial time that $u \in A$ and the same for $v \in B$, but you see this is not the case I said, and when I tried to explain why this is wrong, I mumbled and stumbled because I suddenly came to realize that I don't really comprehend the concept to its full depth!
Any way I said if $\langle u,v \rangle$ was not NP-hard to decide, then also would the language $A$ for instance, because you can look at $A\times A$ as a reduction from $A$.
My question is, if I'm right what's my colleague's error and how to explain it, and if I'm wrong which I guess I'm not (this far I do know), then what's wrong about my theory?