This is side content linked from Part I of posts about various definitions of embedded submanifolds of \(\Rd\).
Reminder: the definitions
The main post lists three equivalent definitions of embedded submanifolds of \(\Rd\). It provides a proof that any connected set which conforms to Definition 1 also conforms to Definition 2. From there, it is an easy step to deduce that it also conforms with Definition 3. Just in case you might prefer to go from Definition 1 to Definition 3 directly, check out the section below.
Definition 1 (as a retract) A connected subset \(M \subseteq \Rd\) is an embedded submanifold of \(\Rd\) if there exist an open set \(U \subseteq \Rd\) and a smooth map \(\pi \colon U \to U\) such that:
- \(M = \pi(U)\), and
- \(\pi(x) = x\) for all \(x \in M\).
Definition 2 (via diffeomorphisms) A subset \(M \subseteq \Rd\) is an embedded submanifold of \(\Rd\) if there exists an integer \(r \geq 0\) with the following property:
For all \(x \in M\) there exist:
- A neighborhood \(U\) of \(x\) in \(\Rd\),
- An open set \(V\) in \(\Rd\), and
- A diffeomorphism \(\psi \colon U \to V\)
such that \(\psi(M \cap U) = \{ z \in V : z_{r+1} = \cdots = z_d = 0 \}\).
Definition 3 (as level sets) A subset \(M \subseteq \Rd\) is an embedded submanifold of \(\Rd\) if either (a) \(M\) is open in \(\Rd\), or (b) there exists an integer \(k \geq 1\) with the following property:
For all \(x \in M\) there exist a neighborhood \(U\) of \(x\) in \(\Rd\) and a smooth map \(h \colon U \to \Rk\) such that
- \(M \cap U = \{y \in U : h(y) = 0\}\), and
- \(\D h(x)\) has rank \(k\).
Another proof that the image of \(\pi\) is smooth
It is possible to connect Definition 1 to Definition 3 more directly. All considered, I find it less appealing, but here it is for the record.
The construction here is from nLab, where it is stated with broader context.
Say \(M \subseteq \Rd\) is connected and we have a smooth map \(\pi \colon U \to U\) such that \(M = \pi(U)\) and \(\pi(x) = x\) for all \(x \in M\), as per Definition 1. We want to show that \(M\) is an embedded submanifold of \(\Rd\) as per Definition 3.
From the main post, we know that \(h(y) := y - \pi(y)\) is not a viable local defining function for \(M\).
An alternative is to proceed locally, and to try and extract from the fixed-point condition \(\pi(x) = x\) a subset of just the right number of (independent) equations. To this end, it is useful to tap into our readily available intuition about the dimension of \(M\).
To express the condition that \(y\) is in \(M\), we should require \(y - \pi(y) = 0\). However, this is \(d\) equations in \(d\) variables, which is too many. Really, we only need \(d - r\) equations, where \(r\) is the dimension of \(M\).
Accordingly, split the vector \(y - \pi(y)\) into two parts (recalling that \(\D\pi(x)\) is a projection for each \(x\) in \(M\), and that its rank is the dimension of \(M\)): \[ y - \pi(y) = \D\pi(\pi(y))[y - \pi(y)] + \big( I - \D\pi(\pi(y)) \big)[y - \pi(y)]. \] Looking at the right-hand side, the first term is “along” \(M\) (tangent to \(M\) at \(\pi(y)\)), while the second term is “away from” \(M\) (in the kernel of \(\D\pi(\pi(y))\)).
Both parts must be zero for \(y\) to be in \(M\). The second one—on its own—accounts for the right number of equations, namely, the codimension of \(M\). Thus, it makes sense to aim for a local defining function built out of that second term.
To avoid unnecessary moving parts, let us “fix” the projector \(I - \D\pi(\pi(y))\). To this end, fix a base point \(\bar{x} \in M\). Consider a neighborhood \(V \subseteq U\) of \(\bar{x}\) in \(\Rd\) (arbitrary for now, but we will fix one momentarily). Then, define the projector \[ P_{\bar x} := I - \D\pi(\bar{x}) \] from \(\Rd\) to \(\ker \D\pi(\bar{x})\) (a “normal space” to \(M\) at \(\bar{x}\)). This projection is surjective: it has rank \(k := d - \rank\,\D\pi(\bar{x})\).
We use the projector to define a tentative local defining function \(h\) for \(M\) near \(\bar{x}\), and an associated level set \(Z\): \[ Z := \{ y \in V : h(y) = 0 \} \quad \textrm{ with } \quad h(y) = P_{\bar x}(y - \pi(y)). \] What can we say about \(Z\)?
It is clear that \(M \cap V\) is included in \(Z\), because if \(y\) is in \(M\) then \(y - \pi(y) = 0\).
Moreover, \(Z\) is an embedded submanifold of \(\Rd\) because \(h\) is a local defining function for \(Z\) (Definition 3). To see this, first note that, for all \(y\), \[ \D h(y) = P_{\bar x}(I - \D\pi(y)). \] Then, looking at \(y = \bar{x}\) specifically, we find \(\D h(\bar{x}) = P_{\bar{x}}^2 = P_{\bar{x}}\): this is surjective, as required. Just make \(V\) small enough so that \(\D h(y)\) remains surjective for all \(y \in V\) (doable since the rank cannot drop suddenly).
What is more, the tangent space to \(Z\) at \(\bar{x}\) is what we expect it to be for \(M\). Indeed, \[\T_{\bar{x}} Z = \ker \D h(\bar{x}) = \im(I - P_{\bar{x}}) = \im \D \pi(\bar{x}),\] where we used the fact that \(P_{\bar{x}}\) is a projector.
This all hints to the resolution: if we can show that \(M\) is not only included in \(Z\) but that the two actually coincide (locally around \(\bar{x}\)), then we will have found that \(h\) is a local defining function for \(M\) too, and hence that \(M\) is an embedded submanifold of \(\Rd\).
Therefore, the only remaining task is as follows:
After possibly shrinking \(V\) some more, show that if \(h(y) = P_{\bar x}(y - \pi(y)) = 0\) (i.e., \(y \in Z\)), then in fact \(y - \pi(y) = 0\) (i.e., \(y \in M\)).
The proof on the nLab page achieves this as follows.
Consider \(\pi^{-1}(V) \subseteq U\): this is a neighborhood of \(\bar{x}\) because \(\pi\) is continuous and \(\pi(\bar{x}) = \bar{x}\). Therefore, \(V' := V \cap \pi^{-1}(V)\) is also a neighborhood of \(\bar{x}\). Since \(Z\) is an embedded submanifold of \(\Rd\), so is the subset \[ Z' := Z \cap V'. \] Moreover, for all \(y\) in \(Z'\) we see that \(\pi(y)\) is in \(Z\). Indeed, for \(y \in Z'\) we know \(y\) is in \(V'\) and hence \(\pi(y)\) is in \(V\) by design; and of course we also have \(h(\pi(y)) = 0\) because \(\pi(\pi(y)) = \pi(y)\).
Based on these considerations, restrict \(\pi\) to \(Z'\) to get a smooth map \[ \hat \pi := \pi|_{Z'} \colon Z' \to Z. \] Of course, \(\hat\pi(\bar{x}) = \bar{x}\). Importantly, the differential \(\D \hat \pi(\bar{x}) \colon \T_{\bar{x}}Z' \to \T_{\bar{x}}Z\) is invertible. This is because both tangent spaces coincide with \(\im \D\pi(\bar{x})\) and we know that \(\D \pi(\bar{x})\) acts as the identity on that space.
Therefore, the Inverse Function Theorem applies: it provides neighborhoods \(W' \subseteq Z'\) of \(\bar{x}\) and \(W \subseteq Z\) of \(\bar{x}\) such that \(\hat \pi\) is a diffeomorphism from \(W'\) to \(W\). In particular, there exists an inverse map \(\hat \pi^{-1} \colon W \to W'\) which is also smooth.
The magic happens now. Take any \(y\) in \(W \subseteq Z\). Then, \[ y = \hat \pi(\hat \pi^{-1}(y)) = \pi(\hat \pi^{-1}(y)) \] and so \[ \pi(y) = \pi(\pi(\hat \pi^{-1}(y))) = \pi(\hat \pi^{-1}(y)) = y. \] Therefore, \(y\) is in \(M\).
Stated differently: locally, \(M\) coincides with \(Z\), and we have a local defining function for \(Z\), so we have a local defining function for \(M\).