Is the unweighted vertex cover problem equivalent to its weighted version? Consider the unweighted and weighted versions of the vertex cover problem (UVC and WVC for short, respectively). As UVC is a special case of WVC, is it true that $$\text{UVC} \leq_\mathrm{m} \text{WVC}$$ but $$\text{UVC} \not \equiv_\mathrm{m} \text{WVC} \enspace ?$$

I think the first part is easy to prove, since one can construct an instance of WVC by taking an instance of UVC and letting the weight of all of its vertices be the same.

I find not that easy to prove the second part though. As WVC is a generalization of UVC, intuitively $\text{WVC} \leq_\mathrm{m} \text{UVC}$ can't hold, but I'd appreciate some help to demonstrate it in a formal way.
If you're talking about determining whether there exists a cover of weight $\le T$, then the problem is obviously in NP and thus reduces to unweighted vc (since $VC$ is NP complete). 

If however were talking about whether the minimal cover is of weight $\le T$, then the problem is $\Sigma_2$ complete, and is therefore not reducible to vc unless the hierarchy collapses to the first level.