# Idempotents split in category of smooth manifolds?

In his paper "Qualitative Distinctions Between Some Toposes of Generalized Graphs" (reproduced here), page 267, Lawvere says that the idempotent-splitting completion of the category of open sets of Euclidean spaces $\mathbb{R}^n$ and smooth maps between them is the category of smooth (second-countable, Hausdorff) manifolds and smooth maps.

It's not very hard to see that every smooth manifold is a smooth retract of such an open set, so that the category of smooth manifolds embeds fully and faithfully in the idempotent-splitting completion. However, I haven't been able to find a proof of the converse, that every smooth retract of an open set is in fact a smooth manifold. Equivalently: if $p: U \to U$ is a smooth map on an open set $U$ of $\mathbb{R}^n$ such that $p \circ p = p$, then the fixed-point set $Fix(p) = \{x \in U: p(x) = x\}$ admits a smooth structure so that the inclusion $Fix(p) \hookrightarrow U$ is a smooth embedding. Can someone provide details of a proof?

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Since $p$ is a projection on tangent spaces it can probably be realized (locally) as a standard projection on $\mathbb R^n$. This may lead you to find euclidean neighbourhoods for the image of $p$. –  Fernando Muro Apr 5 '14 at 20:58
@FernandoMuro Yes, that idea certainly occurred to me. But converting that into a detailed proof was causing me some difficulty. –  Todd Trimble Apr 5 '14 at 21:01
Are you sure you don't need second countability in place of paracompactness? A disjoint union of $2^{2^{\aleph_0}}$ copies of any given closed smooth manifold is paracompact (because it is metrizable) but not second countable. It doesn't seem like this is the retract of anything in the category of open sets in Euclidean space, since it doesn't embed in any $\mathbb R^N$ simply for reasons of cardinality. –  John Pardon Apr 5 '14 at 23:44
@JohnPardon Sorry, you're right, I meant second-countable. Good catch. –  Todd Trimble Apr 5 '14 at 23:51

Assume that:

$$0\in Fix(p)$$

and $U$ is a small neighborhood of $0$.

Let $$f(x)=x-p(x)$$

Identifying $\mathbb R^n$ with $T_0\mathbb R^n$, set $\pi_{\mathrm{ker}(dp(0))}$ to be the projection $\mathbb R^n\to\mathrm{ker}(dp(0))$.

Then $g=\pi_{\mathrm{ker}(dp(0))}\circ f$ is smooth and regular at $0$, so $M=g^{-1}(0)$ is a manifold of dimension equal to the rank of $dp(0)$.

By construction, $M$ contains $Fix(p)$.

On the other hand, since $Fix(p)$ is the image of $p$, $p$ preserves $M$.

Thus, $p|_M$ is a smooth retraction of full rank on $M$. In particular, it's a local diffeomorphism (since it has full rank), and since it's a retraction and we're working in a small neighborhood of $0$, it's the identity.

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I'm sorry, I'm having some trouble understanding this. For starters, what is $\pi_{\ker(dp(0))}$? –  Todd Trimble Apr 5 '14 at 22:13
The projection to the kernel of $dp(0)$, after identifying $T_0\mathbb{R}^n$ with $\mathbb{R}^n$ itself. –  Zack Apr 5 '14 at 22:18
Thanks for adding more details. I've looked this over pretty carefully now, and your argument seems to be holding up. So acceptance and another thank-you is in order! –  Todd Trimble Apr 5 '14 at 23:39
Nice proof. I wonder if in fact it can be squeezed into something like "every such $p$ is locally diffeomorphic to ${\mathbb R}^{k+l}\twoheadrightarrow{\mathbb R}^k\hookrightarrow{\mathbb R}^{k+l}$, $(u,v)\mapsto u\mapsto(u,0)$" - this would make it completely transparent... –  მამუკა ჯიბლაძე Apr 6 '14 at 8:29