Reducing problem A to problem B means that problem B is at least as hard as A, if not even more so.
If I can reduce sorting to some other problem X, I know that X has a lower-bound of Omega(n log n). Is that lower-bound guaranteed to be a tight lower-bound? I suspect that it shouldn't be, because X is only known to be at least as hard as A -- implying that it could be harder, and have a different lower-bound as a result.
I mean that in the sense that it's correct to say that because insertion sort has a worst-case tight upper-bound of O(n^2), it's also correct to say it has a worst-case running time of O(n^3). It's correct, but not of much practical value -- because we're interested in tight bounds 99% of the time.