Suppose we have two separable Hilbert spaces $\mathbb{H}_{1},\mathbb{H}_{2}$ and the compact operator $\mathscr{T}:\mathbb{H}_{1}\to\mathbb{H}_{2}$.
We know that since $\mathscr{T}$ is compact, its inverse does not exist. What if we consider though the operator $\tilde{\mathscr{T}}:\overline{Im(\mathscr{T}})\to Im(\mathscr{T})$ that maps $x\mapsto\mathscr{T}x$. Then, this operator is invertible. Is it also bounded?
Your $\tilde{\mathscr T}$ need not be invertible. For example, if $\{u_n\}_{n=0,1,2,\ldots}$ is an orthonormal basis of $\mathbb H = \mathbb H_1 = \mathbb H_2$, define $\mathscr T$ by $$ \eqalign{\mathscr T u_0 &= 0\cr \mathscr T u_n &= \frac{u_{n-1}}{n} \ \text{for}\ n \ge 1 \cr}$$ Note that $\text{Im}(\mathscr T)$ is dense in $\mathbb H$ (e.g. it contains all finite linear combinations of the basis vectors).
If $\text{Im}(\mathscr T)$ is infinite-dimensional, $\tilde{\mathscr T}$ can't have a bounded inverse. That's a consequence of the fact that the unit ball in an infinite-dimensional Hilbert space is not compact.