Let $A$ and $B$ be C*-algebras, let $I\subseteq A$ and $J\subseteq B$ be closed, two-sided ideals, and let $\pi\colon A\to B$ be a bijective, linear map satisfying $\pi(I)=J$. Assume that both the restriction of $\pi$ to $I$ and the induced map $A/I\to B/J$ are continuous.
Is $\pi$ continuous?
No.
Counterexample. Let $C$ be an infinite-dimensional $C^*$-algebra and let $A = B = C \oplus C$. We set $I = J = C \oplus \{0\}$.
Let $\varphi: C \to C$ be a non-continuous linear mapping and define $\pi: A \to A$ by means of the operator matrix $$ \begin{pmatrix} \operatorname{id} & \varphi \\ 0 & \operatorname{id} \end{pmatrix}. $$ Then $\pi$ is bijective and leaves $I$ invariant. It's restriction to $I$ is the identity operator (in particular, $\pi(I) = I$), and the induced operator $A/I \to A/I$ is the identity operator, too. However, $\pi$ is not continuous.
