# Continuous real function on germs

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?

## 1 Answer

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 $[g] \in U$.)

• 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. – Igor Khavkine Aug 10 '14 at 0:35
• Sure, no problem. – Nik Weaver Aug 10 '14 at 3:14