Let $X$ be an integral scheme of finite type over a field. Then there is a surjective finite map $\tilde{X} \to X$ from the normalization $\tilde{X}$ of $X$. 


Is this going to be bijective?


In the simplest non-normal case, namely the spectrum of $k[x^2, x^3] = k[t,u]/(t^3 -u^2)$, the map is bijective, because the curve is geometrically just a cubic curve (the set of all $(v^2, v^3)$ in affine 2-space, geometrically). I can't find it asserted anywhere that the map is necessarily bijective, though. 



Is the normalization map bijective?
Your example is not really the simplest case, in that a cusp is not the simplest possible curve singularity. Rather, the simplest curve singularity is a node, e.g. as in 
$y^2 = x^3 - x^2$. There are then two branches passing through the singularity, and
the normalization map separates them. (Exercise: The normalization is again given by a map
$t \mapsto x(t),y(t)$; find this map!)


This is the typical phenomenon with normalization: it separates the different branches passing through a singularity. The situation of having just a single branch (and hence having a bijective normalization map) is somewhat unusual (and has a special adjective to describe it: *unibranch*).


