Intuitively, a (smooth) embedded submanifold of \(\Rd\) is a subset \(M \subseteq \Rd\) which is “smooth” in the sense that, locally, “it can be smoothly deformed to a linear subspace, and back.” Think of a sphere or the shell of a doughnut, but not a cube or a cross.
We can make this concrete in several equivalent ways. The definition we choose will affect everything else we do (from defining subordinate concepts to proving theorems about them), thus we should choose wisely.
This blog post is about a classical yet little known definition. Notice how little prior background it requires: any student who took calculus can take this in.
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 \(x \in M\).
For example, this definition deems (correctly) that the unit sphere is an embedded submanifold of \(\Rd\) because \(\pi(x) := x/\|x\|\) is an appropriate choice with \(U = \Rd \backslash \{0\}\).
Definition 1 is not standard, yet it is equivalent to all the ones you may have seen (see below): we prove as much in this post.
I became aware of this fact via a book of Michor (2008), where it is stated as Theorem 1.15. It also appears in a book by Hirsch (1976) as a starred exercise (namely, Ex. 2 in Sec. 1.2). This topic is further discussed in a post on Math Overflow, which cites a 1987 paper by Lawvere (p. 267) and links to a proof on nLab.
My motivation for writing this is that, if you need to teach someone what a manifold is from scratch, or if you need to build up some definitions as brief background for a paper, this is a credible option: see Part II.
Quick comments
Throughout, “smooth” means \(C^\infty\). It all generalizes to \(C^p\).
In topology, a map \(\pi\) as in Definition 1 is called a (topological) retraction, and \(M\) is a retract of \(U\) through \(\pi\). Given that the word “retraction” refers to a different concept in optimization on manifolds, we might call \(\pi\) a defining retraction for example.
One subtlety is that the definition requires \(M\) to be connected. If it is not, then apply the definition to each connected component separately. If they all are embedded submanifolds of the same dimension, then \(M\) as a whole is an embedded submanifold. The dimension of \(M\) will be defined as we go—think: rank of \(\D\pi(x)\) at any \(x \in M\) (it happens to be constant).
The map \(\pi\) is idempotent and \(M\) is its fixed points
Consider a map \(\pi \colon U \to U\) as in Definition 1. In particular, if \(x\) is in \(M\), then \(\pi(x) = x\). The other way around, if \(\pi(y) = y\) (for any \(y \in U\)) then \(y\) is in the image of \(\pi\), which is \(M := \pi(U)\). Thus, \[ M = \{ y \in U : \pi(y) = y \}. \] In words: \(M\) is the set of fixed points of \(\pi\).
Moreover, \(\pi\) is idempotent because for all \(y \in U\) it holds that \(\pi(y)\) is in \(M\) so \(\pi(\pi(y)) = \pi(y)\). Succinctly: \[ \pi \circ \pi = \pi. \] In fact, we could even have this instead of the second condition in Definition 1. Indeed, if \(\pi \circ \pi = \pi\) and \(M := \pi(U)\), then each \(x \in M\) is of the form \(x = \pi(y)\) for some \(y \in U\) so \(\pi(x) = \pi(\pi(y)) = \pi(y) = x\).
Thus, the main message of this post can be summarized as:
The connected embedded submanifolds of \(\Rd\) are the images of smooth idempotent maps.
Contrast to more classical definitions
The intrinsic way to define submanifolds is to define first what a manifold is (through charts and atlases), then to define submanifolds as follows (see the textbook by Lee (2012), on page 98 (!)):
An embedded submanifold of \(\Rd\) is a subset \(M \subseteq \Rd\) that is a manifold in the subspace topology, endowed with a smooth structure with respect to which the inclusion map \(i \colon M \to \Rd \colon x \mapsto i(x) = x\) is a smooth embedding.
This is a mouthful; it’s not particularly illuminating; and most importantly: it requires one to first ascertain that \(M\) is a manifold in its own right (thus, with charts and atlases) before moving on to checking all the other properties that require their own definitions. Somehow, this is the standard definition, but it’s hardly a good entry point.
The next definition more closely parallels the intuition conveyed by the first sentence of this post. I like it because it convincingly “captures” the idea of smoothness. Explicitly, it requires that, locally around each \(x\), the space \(\Rd\) can be deformed smoothly in a way that a patch of \(M\) around \(x\) (namely, \(M \cap U\)) becomes flat; and that deformation can be undone smoothly as well.
The fact that this is equivalent to the former definition is standard (Lee 2012, Prop. 5.8).
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 \(\psi(M \cap U) = \{ z \in V : z_{r+1} = \cdots = z_d = 0 \}\).
Yet another approach is to require that \(M\) is locally the zero-set of a smooth function together with a (truly necessary) rank condition. Here too, it is well known that this is equivalent to the definition above (Lee 2012, Prop. 5.16) (via the Inverse Function Theorem), hence we can also use this one as the entry point to differential geometry.
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 \(k\).
One advantage of this definition is that manifolds often come up naturally as level sets of functions \(h\) (called local defining functions). Think for example of a sphere, defined by \(h(x) := x^\top x - 1 = 0\).
Both Definition 2 and Definition 3 are still a mouthful, but at least they are phrased entirely using terms that will be familiar to anyone who took a first course in calculus and linear algebra.
In spite of these qualities, a somewhat annoying aspect of both definitions is that, in general, \(M\) cannot be defined with a a single diffeomorphism (because if it can be, then it must be diffeomorphic to \(\Rm\)) nor with a single local defining function \(h\) (because if it can be, then it must be orientable (Lee 2012, Prop. 15.23)).
In contrast, Definition 1 describes the entire manifold \(M\) with a single, nice map \(\pi\). Downstream, this notably has the effect that further definitions and proofs need not be articulated as “for each \(x\), consider a \(\psi\) or an \(h\) etc.”: we can process the whole manifold in one shot.
There is a \(\pi\) for each submanifold
If \(M\) is an embedded submanifold of \(\Rd\) as per one of the standard definitions, then we can build a map \(\pi\) that conforms to Definition 1 in two (related) ways:
Via the Tubular Neighborhood Theorem (TNT): see (Lee 2012, Prop. 6.25).
Via metric projection: let \(\pi(y) := \arg\min_{x \in M} \|x - y\|^2\), with sufficiently small domain \(U\) (a neighborhood of \(M\)) so that \(\pi\) is well defined and smooth.
That second construction has the benefit of being actionable (for example, \(\pi(y) = y/\|y\|\) is exactly that map for the sphere, and it is easily figured out). As it turns out, the fact that metric projection is (locally) smooth “falls out of” the proof of the TNT, so they are really one and the same (Lee 2012, Pb. 6-5).
Accordingly, to confirm that Definition 1 is equivalent to the standard definitions, we only need the other direction: given a defining retraction \(\pi\), show that \(M\) is an embedded submanifold in the standard sense. The next three sections do exactly that.
The differential of \(\pi\) is a projector
Since \(\pi \circ \pi = \pi\), for all \(y \in U\) we have by the chain rule: \[ \D \pi(\pi(y)) \circ \D \pi(y) = \D \pi(y). \] In particular, for all \(x \in M\) we have \(\pi(x) = x\) so that \[ \D \pi(x) \circ \D \pi(x) = \D \pi(x). \] Thus, \(\D \pi(x)\) is an idempotent linear map: a projector.
As a side question to build intuition, we may ask: what does it project onto?
Well, if \(v\) is in the image of \(\D\pi(x)\), then there exists \(w\) such that \(v = \D \pi(x)[w]\). Let \(c(t) = \pi(x + t w)\): this is a smooth curve in \(M\) which passes through \(x\) at \(t = 0\) with velocity \(c'(0) = \D \pi(x)[w] = v\).
Conversely, if \(v\) is the velocity at \(t = 0\) of a smooth curve \(c\) in \(M\) with \(c(0) = x\), then \(c(t) = \pi(c(t))\) for all \(t\) (because \(c(t)\) is in \(M\)), so \[ v = c'(0) = \D \pi(c(0))[c'(0)] = \D\pi(x)[v]. \]
Combined, we found that:
The image of \(\D \pi(x)\) is the set of velocities of all smooth curves in \(M\) as they pass through \(x\).
For this reason:
In Part II, we shall call \(\im \D\pi(x)\) the tangent space to \(M\) at \(x\). Accordingly, the dimension of \(M\) at \(x\) will be defined as the rank of \(\D \pi(x)\), that is, the dimension of that tangent space.
In the next section, we see that \(M\) has the same dimension at each point \(x\), hence we shall simply call this the dimension of \(M\).
The map \(\pi\) has constant rank
Any map \(\pi\) that conforms to Definition 1 has a differential of constant rank, at least in a neighborhood \(M\). More precisely, the following is true:
Lemma 1 Let \(\pi \colon U \to U\) be smooth and idempotent (\(\pi \circ \pi = \pi\)). Then, there exists a neighborhood \(V \subseteq U\) of \(\pi(U)\) such that \(y \mapsto \rank\,\D\pi(y)\) is constant in each connected component of \(V\).
Proof. The argument here exactly tracks the proof in (Michor 2008, Thm. 1.15).
Let \(M = \pi(U)\). Then:
- For all \(x \in M\), we have \(\ker(I - \D\pi(x)) = \im \D\pi(x)\).
Indeed, recall \(A := \D\pi(x)\) is a projector (\(A^2 = A\)). On the one hand, for \(u \in \ker(I-A)\) we have \(u = Au\) hence \(u \in \im A\). On the other hand, for \(u \in \im A\) there exists \(v\) such that \(u = Av\) hence also \(Au = A^2 v = Av = u\), confirming \(u \in \ker(I-A)\).
- Therefore, for all \(x \in M\) we have from the rank-nullity theorem that \[ d = \dim \ker(I - \D\pi(x)) + \dim \im(I - \D\pi(x)) = \rank\,\D\pi(x) + \rank(I - \D\pi(x)). \]
Neither of these ranks can suddenly drop, yet their sum is constant. Thus, they must both be constant on each connected component of \(M\). For convenience, assume \(M\) is connected (the argument easily extends). Let \(r\) be the rank of \(\D\pi(x)\) for all \(x \in M\).
- The rank of \(\D\pi(y)\) is at most \(r\) for all \(y \in U\).
The follows from idempotence, as then for all \(y\) we have \(\D\pi(y) = \D(\pi \circ \pi)(y) = \D\pi(\pi(y)) \circ \D\pi(y)\). Since \(\pi(y)\) is in \(M\), we see that \(\D\pi(\pi(y))\) has rank \(r\), and so \(\D\pi(y)\) must have rank at most \(r\).
- The rank of \(\D\pi(y)\) is at least \(r\) for all \(y\) in some neighborhood $V $ of \(M\).
This is because \(\rank\,\D\pi(x) = r\) for all \(x \in M\) and (again) the rank cannot suddenly drop: it can only stay constant or (in general) suddenly increase.
- Combined, the last two claims show that \(\rank\,\D\pi(y) = r\) for all \(y \in V\).
The proof is complete.
The lemma notably implies that \(\im \D\pi(y) = \im \D\pi(\pi(y))\) for all \(y \in V\). Moreover, combined with the fact that \(M := \pi(U)\) is a smooth manifold of dimension \(r\) (which we prove in the next section), the lemma also implies that \[ \pi \colon V \to M \] is a smooth submersion, an open map and a quotient map (Lee 2012, Prop. 4.28). (But we cannot use this yet.)
The image of \(\pi\) is smooth
The argument here corresponds to the in-between-the-lines part of the proof in (Michor 2008, Thm. 1.15), as per an exchange with Peter Michor on Math Overflow.
Let \(\pi \colon U \to U\) be smooth and idempotent such that \(M := \pi(U)\) is connected. If need be, use Lemma 1 to replace \(U\) with a smaller neighborhood of \(M\) such that \(\D\pi(y) =: r\) for all \(y \in U\).
The goal is to show \(M\) is an embedded submanifold of \(\Rd\) “in the usual sense”. Specifically, we target Definition 2 (because it is so close to the intuitive notion of smoothness).
Fix a point \(x \in M\). Notice \(\pi(x) = x\). By the Constant Rank Theorem (Lee 2012, Thm. 4.12), there exist:
- Two neighborhoods \(V, W\) of \(x\) in \(U\) such that \(\pi(V) \subseteq W\), and
- Two diffeomorphisms \(\varphi \colon V \to \Rd\) and \(\psi \colon W \to \Rd\)
such that \[ \tilde\pi := \psi \circ \pi \circ \varphi^{-1} \colon \Rd \to \Rd \] is the linear projection \(\tilde\pi(z_1, \ldots, z_r, z_{r+1}, \ldots, z_d) = (z_1, \ldots, z_r, 0, \ldots, 0)\).
The image of \(\tilde \pi\) is a linear subspace of \(\Rd\) of dimension \(r\). Consider the pre-image of that subspace through \(\psi\): \[ Z := \psi^{-1}(\im \tilde\pi) = \{ y \in W : \psi(y) \in \im \tilde\pi \} = \{ y \in W : \psi(y) = (\tilde\pi \circ \psi)(y) \}. \] (It is already clear from Definition 2 that \(Z\) is an embedded submanifold of \(\Rd\), of dimension \(r\).) For convenience, restrict it to \[ Z' := Z \cap V, \] so that \(Z'\) is included in \(V \cap W\). We aim to show that this is a “patch of \(M\)” around \(x\).
For all \(y\) in \(V\), notice that \[ \psi(\pi(y)) = (\tilde\pi \circ \varphi)(y). \] Here comes the key step: since \(\tilde\pi \circ \tilde\pi = \tilde\pi\), it follows that \[ (\tilde\pi \circ \psi)(\pi(y)) = (\tilde\pi \circ \varphi)(y) = \psi(\pi(y)). \]
From here, we see two things:
If \(y\) is in \(M \cap (V \cap W)\), then \(\pi(y) = y\) so that \((\tilde\pi \circ \psi)(y) = \psi(y)\), that is, \(y\) is in \(Z'\).
If \(y\) is in \(Z'\), then \(\psi(y) = (\tilde\pi \circ \psi)(y)\) so that \(y = (\psi^{-1} \circ \tilde\pi \circ \psi)(y) = (\pi \circ \varphi^{-1} \circ \psi)(y)\). Thus, \(y\) is in the image of \(\pi\), which is \(M\).
Therefore, \(M \cap (V \cap W) = Z'\), and \(\psi|_{V \cap W}\) is a diffeomorphism that “straightens out” the patch \(M \cap (V \cap W)\). This is true around each \(x \in M\), thus \(M\) is an embedded submanifold of \(\Rd\) in the sense of Definition 2.
Could we have built a single defining function \(h\)?
We argued early on that \(M = \{y \in U : \pi(y) = y\}\). Thus, it is tempting to define \[ h(y) := y - \pi(y) \] and to try to show that this is a defining function for \(M\), connecting to Definition 3 (though in a more general sense where \(h\) need not have maximal rank, but only constant rank in a neighborhood of \(M\), as per the Constant Rank Theorem, see (Lee 2012, Thm. 5.12)).
Clearly, \(M = h^{-1}(0)\), so it is just a matter of checking that \(\D h\) has constant rank on a neighborhood of \(M\). What is more, we saw in Lemma 1 that \(\D\pi\) has constant rank on a neighborhood of \(M\), so perhaps \(\D h\) does too? Unfortunately, it does not.
To see clearly why that is, consider the example of a sphere \(\Sd \subset \Rd\):
- \(U = \Rd \backslash \{0\}\).
- \(\pi(y) = y/\|y\|\) for all \(y \in U\).
- \(h(y) = y - \pi(y) = y - y/\|y\|\) for all \(y \in U\).
The differential of \(h\) at \(y\) is \[ \D h(y) = I - \D \pi(y) = I - \frac{1}{\|y\|}I + \frac{1}{\|y\|^3} y y^\top. \] As expected, if \(\|y\| = 1\) (that is, exactly on the sphere), then \(\D h(y) = yy^\top\) has rank \(1\). However, as soon as \(y\) leaves the sphere, the rank of \(\D h(y)\) jumps to \(d\).
Still, one could reasonably ask: if \(M\) is an embedded submanifold of \(\Rd\), is it possible to find a single defining function \(h \colon U \to \Rk\) such that \(M = h^{-1}(0)\) and \(\D h(x)\) has constant rank \(d-r\) for all \(x \in U\)? As noted earlier, it is not always possible to do this with \(k = d-r\) (full rank), because some submanifolds are not orientable. I am not sure if it can be done with some \(k > d-r\) though: please e-mail me if you know.
Another proof that the image of \(\pi\) is smooth
It is also 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 previous section, 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: there is a neighborhood of \(\bar{x}\) in \(Z\) (namely, \(W\)) which is included in \(M\). And we already knew that there is a neighborhood of \(\bar{x}\) in \(M\) (namely, \(M \cap V\)) which is included in \(Z\). Thus, \(M\) and \(Z\) coincide locally around \(\bar{x}\). Since \(Z\) is an embedded submanifold of \(\Rd\), so is \(M\) around \(\bar{x}\). And \(\bar{x}\) was arbitrary in \(M\), so the whole set \(M\) is an embedded submanifold of \(\Rd\).
Building up differential geometry from here
Join me in Part II to build tangent spaces, smooth maps, differentials, local frames, Riemannian metrics, gradients, etc. based on Definition 1.
References
Citation
@online{boumal2026,
author = {Boumal, Nicolas},
title = {A Different Way to Define Embedded Submanifolds: {Part} {I}},
date = {2026-02-10},
url = {www.racetothebottom.xyz/posts/smooth-retracts-are-manifolds-part1/},
langid = {en},
abstract = {There are several ways to define manifolds, and the
definition we choose conditions all the work that follows. One
little known definition may just be the best entry point yet: an
embedded submanifold of \$\textbackslash Rd\$ is nothing but the
image of a smooth idempotent map.}
}