Partial Isometries: Final Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.
Consider an operator:
$$J\in\mathcal{B}(\mathcal{H},\mathcal{K}):\quad P:=J^*J$$
By the C*-property:
$$J=JJ^*J\iff P^2=P=P^*$$
Note that in any case:
$$\mathcal{R}:=\mathcal{N}^\perp:\quad\mathcal{N}:=\mathcal{N}P=\mathcal{N}J$$
By an earlier thread:
$$\|J\varphi\|=\|\varphi\|\quad(\varphi\in\mathcal{R})\iff P\varphi=\varphi\quad(\varphi\in\mathcal{R})$$
It is well-known:
$$P^2=P=P^*\iff P\varphi=\varphi\quad(\varphi\in\mathcal{R})$$

All these together give:
  $$J=JJ^*J\iff\|J\varphi\|=\|\varphi\|\quad(\varphi\in\mathcal{R})$$

Moreover one has:
$$J^*\in\mathcal{B}(\mathcal{K},\mathcal{H}):\quad\mathcal{N}J^*=\mathcal{R}J^\perp$$
Does this admit a direct check?
 A: Assume that $J=JJ^*J$. 
If $\varphi\in\mathcal R=(\ker J)^\perp=\overline{\mathcal R J^*}$, then for any $\xi\in\mathcal H$ we can write $\xi=J^*\eta+\nu$ with $\nu\in(\mathcal RJ^*)^\perp=\ker J$, so $\nu\perp\varphi$ and 
\begin{align}
\langle J^*J\varphi,\xi\rangle=\langle J^*J\varphi,J^*\eta+\nu\rangle=\langle J^*J\varphi,J^*\eta\rangle=\langle JJ^*J\varphi,\eta\rangle=\langle J\varphi,\eta\rangle
=\langle \varphi,J\eta\rangle=\langle\varphi,\xi\rangle.
\end{align}
As $\xi$ was arbitrary, $J^*J\varphi=\varphi$. Now
$$
\|J\varphi\|^2=\langle J\varphi,J\varphi\rangle=\langle J^*J\varphi,\varphi\rangle=\langle \varphi,\varphi\rangle=\|\varphi\|^2.
$$
$$ \ $$

Conversely, if $\|J\varphi\|=\|\varphi\|$ for every $\varphi\in\mathcal R=\overline{\mathcal R J^*}$, we have for any $\xi\in\mathcal H$, writing $\varphi=J^*\xi$,
$$
\langle [(JJ^*)^2-JJ^*]\xi,\xi\rangle=\langle JJ^*\varphi,\varphi\rangle-\langle\varphi,\varphi\rangle=\|J\varphi\|^2-\|\varphi\|^2=0.
$$
As $\xi$ was arbitrary, by polarization we get $(JJ^*)^2=JJ^*$. Now any $\xi\in\mathcal H$ can be written as (a limit of) $\xi=J^*\eta+\nu$, with $\nu\in(\mathcal R J^*)^\perp=\ker J$. So
$$
JJ^*J\xi=JJ^*JJ^*\eta+JJ^*J\nu=JJ^*\eta=J\xi.
$$
That is, $JJ^*J=J$. 
