Least Square Fitted Vector Through SVD Equals to y 
Elements of Statistical Learning, p 66
The SVD of the $N \times p$ matrix $X$ has the form
  $X = UDV^T$
Here $U$ and $V$ are $N \times p$ and $p \times p$ orthogonal matrices, with the columns of $U$ spanning the column space of $X$, and the columns of $V$ spanning the row space.
$D$ is a $p \times p$ diagonal matrix, with diagonal entries $d_1 \ge d_2 \ge ... \ge d_p \ge 0$ called the singular values of $X$. If one or more values $d_j = 0$, $X$ is singular.
Using the singular value decomposition we can write the least squares fitted vector as :
$X \hat{\beta_{ls}} = X(^T)^{−1}^Ty = UU^Ty$

The demonstration is available
here p 266, and makes total sense.
Yet, as $U$ is orthogonal, it seems to me that it implies that $UU^Ty=Iy=y$, and thus that 
$X\hat{\beta_{ls}}=y$, which makes not sense. Indeed, it would mean that the the residual vector $y-X\hat{\beta_{ls}}$  is null. But this should only true in the specific case where $y$ can be expressed as a linear combination of the columns of $X$.
It seems to me that i'm totally missing something but i don't know what it is !
 A: $U$ has orthogonal columns but it's not square and isn't an orthogonal matrix. $UU^T \neq I$ in general since it only has rank $p$ and $UU^T$ involves dot products of the rows which don't have to be orthogonal when $p<n$. If $U$ was square then it would be the case that $UU^T = I$ which makes sense since then $\hat y = y$ since $X$ is a bijection. 
I'm updating to add this example of why this isn't possible if $n > p$.
Let
$$
U = \left(\begin{array}{cc} a & b\\ c&d \\ e&f\end{array}\right).
$$
Suppose $U^TU = I_2$ and $UU^T = I_3$. These turn into the following 9 constraints:
$$
a^2+c^2+e^2 = b^2+d^2+f^2 = 1 \hspace{2cm}(1)\\
ab + cd + ef = 0 \\
a^2 + b^2 = c^2 + d^2 = e^2 + f^2 = 1  \hspace{2cm}(2)\\
ac + bd = ae + bf = ce + df = 0
$$
But we only have 6 variables, and as it turns out these are not satisfiable. To see this, note that we have
$$
a^2 + b^2 + c^2 + d^2 + e^2  +f^2 = 2
$$
from adding the constraints in (1) together, but adding the constraints in line (2) together gives
$$
a^2 + b^2 + c^2 + d^2 + e^2  +f^2 = 3
$$
which is impossible. Thus unless $n=p$ we don't have enough freedom to make $U^TU = I$ while also having $UU^T = I$. Since $U^TU = I$ is guaranteed by construction, the result is that $UU^T \neq I$ unless $n=p$.

I think interpreting this can also help with this. I like to interpret $\hat y = UU^Ty$ as follows: we can expand $U$ to an $n\times n$ matrix $\tilde U = (U \mid U_\perp)$ that actually does contain an orthonormal basis for $\mathbb R^n$ (this can be done via Gram-Schmidt). In this basis we'll have $y = \tilde U \tilde z$ where $\tilde z \in \mathbb R^n$ is $y$'s coordinate w.r.t. $\tilde U$. I'll partition $\tilde z = {z \choose z_\perp}$ where $z \in \mathbb R^p$ gives the coordinates for the basis vectors in $U$, and $z_\perp$ analogously gives the coordinates for the basis vectors in $U_\perp$.
This means
$$
y = \tilde U \tilde z = Uz + U_\perp z_\perp
$$
so
$$
U^Ty = U^TUz + U^TU_\perp z_\perp = z
$$
therefore $U^Ty$ picks out the coordinates of $y$ w.r.t. just the basis vectors in $U$. And then $UU^Ty = Uz$ gives the corresponding column space element in $\mathbb R^n$ with just those coordinates. So it's like doing a change of basis but only keeping the coordinates from $U$, and then getting the column space vector specified by those coordinates. 
This reflects the fact that when we're talking about a $p$-dimensional subspace in $\mathbb R^n$, we can think of the actual vectors in $\mathbb R^n$, which are like $X\beta$, or we can think about the coordinates indexing the column space which is like $\beta$. 
A: It says that $U \in \mathbb{R}^{n \times p}$ so the columns are orthogonal but the matrix is not necessarily orthogonal in the sense that $UU^T =I$ ($\dim(UU^T) =p$)
