Is the normalization map bijective? 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. 
 A: No, the map isn't generally bijective.  For curves, normalization is the same as resolution of singularities.  Look at a nodal cubic curve $y^2=x^2(x-1)$ in the plane.  Over the node, the normalization has two points.
A: 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).
A: If $X$ is reduced and finite type over an algebraically closed field of characteristic zero, then there is a largest finite map $Y \to X$ which is bijective (where $Y$ is also reduced).  This map is the seminormalization, $Y = X^{SN}$.  In fact, it always factors the normalization map $X^N \to X^{SN} \to X$ where $X^N$ is the normalization.
