1
$\begingroup$

Let $C_0^{m,n}$ be the space of germs of continuous maps from $\mathbb{R}^m$ to $\mathbb{R}^n$, located at $0\in\mathbb{R}^m$, with the usual inductive limit topology. One can also consider $C_0^{m,n}$ to be the stalk at $0$ of the sheaf of continuous $\mathbb{R}^n$ valued functions on $\mathbb{R}^m$. I'm trying to get a better idea of the topology on this space of germs. So, here's my question: Are there any continuous functions $F\colon C_0^{m,n} \to \mathbb{R}$ other than those that factor through the evaluation map $[f] \mapsto f(0)$?

If there are more, what are some examples? If there aren't, what's the underlying topological reason?

$\endgroup$

1 Answer 1

4
$\begingroup$

Do you mean that you give $C_0^{m,n}$ the quotient topology? Then I think the answer is that there are no other continuous functions, on the grounds that $C_0^{m,n}$ is badly non-Hausdorff: two germs that have the same value at $0$ belong to the same open sets. So factoring out that equivalence is effectively the same as evaluating at $0$.

(The claim is easy. Suppose given any open set $U$ in $C_0^{m,n}$ and $[f] \in U$, $[g] \in C_0^{m,n}$ such that $f(0) = g(0)$. Then $U$ pulls back to an open subset $\widetilde{U}$ of $C(B^m; \mathbb{R}^n)$ where $B^m$ is the closed unit ball in $\mathbb{R}^m$. So there exists $\epsilon > 0$ such that $\|f - h\|_\infty < \epsilon$ implies $h \in \widetilde{U}$. Then we can find a neighborhood of $0$ on which $\|f - g\|_\infty < \epsilon/2$ and find a function that agrees with $g$ on that neighborhood and belongs to $\widetilde{U}$. Thus $[g] \in U$.)

$\endgroup$
2
  • $\begingroup$ Thanks for spelling out the argument! So it's the non-Hausdoffness that is responsible for the scarcity of real continuous functions. That's exactly what I wanted to see. $\endgroup$ Aug 10, 2014 at 0:35
  • $\begingroup$ Sure, no problem. $\endgroup$
    – Nik Weaver
    Aug 10, 2014 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.