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…?
Best Answer
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.