My concrete example is that the determinant is a group homomorphism between $GL(n, \mathbb{C})$ and the non-zero complex numbers. But we know that $GL(n, \mathbb{C}) = M(n, \mathbb{C})^{\times}$ is the multiplicative group of all $n\times n$ linear transformations $M(n, \mathbb{C})$ and the non-zero complex numbers $\mathbb{C}^{\times}$ is the multiplicative group of the complex numbers $\mathbb{C}$. So we have
$\text{det}:M(n, \mathbb{C})^{\times} \rightarrow \mathbb{C}^{\times}$
is a homomorphism. But $\text{det}$, defined or continuously extened over all of $M(n, \mathbb{C})$ also has the property that it maps non-invertible elements of $M(n, \mathbb{C})$ to 0, the non-invertible element of $\mathbb{C}$.
Can this idea be generalized? Suppose $A$ and $B$ are rings and we have a homomorphism $\phi$
$$ \phi:A^{\times} \rightarrow B^{\times} $$
Is it true that we can continuously extend $\phi$ to $A$ by $\tilde{\phi}$ and that when we do so we will find something like $\tilde{\phi}(A-A^{\times})\subset B-B^{\times}$? If not are there some additional conditions that make something like this true?
Maybe topology is required so the notion of continuity makes sense...?
Extension by zero is always possible: the set of non-invertible elements of a ring form a two-sided ideal, so defining $\tilde{\phi}: A\to B$ by letting $\tilde{\phi}|_{A^\times} = \phi$ and $\tilde{\phi}_{A \setminus A^\times} = 0$ is a homomorphism. A more interesting question might be when this is the only such extension. The answer is not always, since the identity map on $A^\times$ can be extended to the identity map on $A$.
It's not always true that in such an extension the non-invertible elements must be sent to non-invertible elements; consider the extension of $\mathbb{Z}^\times \hookrightarrow \mathbb{Q}^\times$ to $\mathbb{Z} \hookrightarrow \mathbb{Q}$, or similar examples involving localization.