# Compact-open and Whitney $C^\infty$-topologies agree on $C^\infty(M, N)$ for compact $M$.

Let $$M$$ and $$N$$ be smooth manifolds. There are different topologies we can equip the space $$C^\infty (M, N)$$ of smooth mappings between them with. Two of them are the compact-open topology and the Whitney $$C^k$$-topologies for $$k \in \mathbb{N} \cup \{\infty\}$$ (I'm using this definition based on jet bundles).

I have read claims that when $$M$$ is compact these agree. Is this true? Could someone provide a proof or a reference?

EDIT: I'm starting to seriously doubt that this is true for $$k \geq 1$$. I can show that the Whitney topology is finer than the compact-open, but for say $$f: \mathbb{S}^1 \to \mathbb{R}$$ allowing to change the value by any $$\varepsilon$$ on a compact set with nonempty interior is sufficient to make the derivative arbitrarily large.

Let $$d$$ be a metric on $$N$$. Then the following family is a local basis of the Whitney topology on $$C(M,N)$$: $$B(f,\varepsilon) := \{g \in C(M,N): d(g(x),f(x)) \lt \varepsilon(x) \text{ for all x} \in M, \varepsilon \in C(M, (0; +\infty))\}$$
Since $$M$$ is compact, we see that for any continuous function $$\varepsilon \colon M \to (0; +\infty)$$ there exist $$n \in \mathbb{N}$$ such that $$\frac{1}{n} \lt \varepsilon(x)$$ for all $$x \in M$$ and compact-open topology and topology of uniform convergence coincide. Hence $$\{B(f, \frac{1}{n}), n \in \mathbb{N}\}$$ is a local basis for both topologies. Therefore compact-open topology and Whitney topology on $$C(M,N)$$ coincide. Compact-open $$C^{k}$$-topology and Whitney $$C^k$$-topology can be defined by embedding $$j^k \colon C^{\infty}(M,N) \to C(M, J^k(M,N))$$. For details see Michor P.W. Manifold of differentiable mappings, page 33. From previous it follows that they coincide too.