4.7Embeddings of Manifolds

Updated 22 Aug 2026Chapter (PDF)

We have shown that regular, second-countable spaces can be embedded into $\bR^\omega$ . A natural question is to ask which type of space $X$ can be embedded into some finite-dimensional Euclidean space $\bR^n$ . We show that compact manifolds satisfy this property.

[4.7.1]Definition(

m

-Manifold)#

A space $X$ is said to be an $m$ -manifold if $X$ is Hausdorff, second-countable, and is locally homeomorphic to $\bR^m$ . That is, for any $x \in X$ , there is some open neighborhood $U$ of $x$ such that $U$ is homeomorphic to some open subset of $\bR^m$ .

[4.7.2]Example(Examples of Manifolds)#

A curve (with some regularity conditions imposed) is a $1$ -manifold, and the same for a surface being a $2$ -manifold.

[4.7.3]Definition(Support of a Map)#

Suppose $\phi: X \to \bR$ is a real-valued map. Then the support of $\phi$ , denoted by $\supp(\phi)$ , is the closure of $\phi^{-1}(\bR \setminus \{0\})$ .

[4.7.4]Remark(Intuition behind Support)#

The support of a function is not simply where the points where the map is nonzero: it's the points where there are no open neighborhoods for which the map vanishes on. We can characterize the support of a function as where it's “active”. A polynomial $p$ may be zero at a root $x = r$ , but it either rebounds or intersects without “wasting any time,” and so we would say that the polynomial is “active” at $r$ , and so $r \in \supp(p)$ . The bottom line is, if a point $x$ is not in $\supp(f)$ , then there is some open neighborhood $U$ of $x$ such that $f|_U \equiv 0$ . Thus the closure makes the support a much better tool to determine where a function is dead. We also note that the support being closed is a useful property (analogously, the complement of the support being open is useful).

[4.7.5]Definition(Partition of Unity)#

Let $\{U_1, \dots, U_n\}$ be some finite covering of the space $X$ . Then a collection of continuous maps $\{\phi_1, \dots, \phi_n\}$ such that $\phi_k: X \to [0, 1]$ is said to be a partition of unity dominated by $\{U_k\}$ if it satisfies the following conditions.

$\supp(\phi_k) \subseteq U_k$ for each $k \in \{1, \dots, n\}$ .
For any $x \in X$ , $\sum_{k = 1}^n \phi_k(x) = 1$ .

[4.7.6]Remark(Intuition behind Partition of Unity)#

Suppose $X$ has a finite cover $\{U_1, \dots, U_n\}$ together with a collection of continuous local maps $\{g_1, \dots, g_n\}$ where each $g_k: U_k \to \bR$ . A natural question is: how can we produce a global map $g: X \to \bR$ that represents these local maps best? There is no guarantee that $g_j$ and $g_k$ agree on the overlap $U_j \cap U_k$ , so using the pasting lemma is not an option.

A naive approach would be to use characteristic functions $\chi_{U_k}$ to toggle the maps on and off, defining $g(x) = \sum_{k=1}^n \chi_{U_k}(x) g_k(x)$ (where the product is taken to be $0$ outside $U_k$ ). However, this fails for two reasons. First, the binary nature of the characteristic function causes discontinuities at the boundaries. If a sequence of points inside $U_k$ approaches a boundary point $p \in \partial U_k$ , the term $\chi_{U_k}(x)g_k(x)$ approaches $g_k(p)$ , which is generally nonzero. But for points immediately outside the boundary, the term is exactly $0$ . Unless the local map $g_k$ happens to naturally vanish at the boundary of $U_k$ , this creates a jump discontinuity. Second, in overlapping regions where multiple sets intersect, the values are blindly added together. If $x \in U_j \cap U_k$ , the sum yields $g_j(x) + g_k(x)$ , double-counting the outputs and artificially inflating the result.

A partition of unity solves this by acting as a continuous approximation of these characteristic functions. Rather than assigning a rigid binary $1$ or $0$ to dictate which open set $x$ belongs to, $\phi_k(x)$ assigns a “fractional ownership” to $U_k$ . Deep inside $U_k$ , away from other sets in the cover, $\phi_k(x) = 1$ and acts exactly like a standard indicator function. In an overlap $U_j \cap U_k$ , however, the functions provide a continuous interpolation that smoothly crossfades between the patches. For instance, suppose an overlap $V = U_j \cap U_k$ intersects no other sets $U_i$ ; by the first condition, all other maps $\phi_i$ must vanish identically on $V$ . For any $v \in V$ , the value $\phi_j(v)$ measures how much “control” $U_j$ exerts over $v$ , and similarly for $\phi_k(v)$ . The second condition then forces $\phi_j(v) + \phi_k(v) = 1$ , ensuring that the total ownership of the point is exactly 100%. Simply put, each map $\phi_k$ determines the amount of control $U_k$ exerts over any point $x \in X$ , with the restriction that $U_k$ claims no ownership outside its boundary and that the the ownership each map exerts over $x$ always sums perfectly to $1$ .

We define the global map as $g(x) = \sum_{k=1}^n \phi_k(x) g_k(x)$ , a convex combination of the local maps with the maps from the partition of unity. Since $\sum \phi_k(x) = 1$ , this behaves as a true weighted average. If the local maps $g_j$ and $g_k$ both output a value of $c$ at an overlap, the global map outputs $0.7c + 0.3c = c$ , which successfully blends conflicting local maps without inflating or deflating the values.

We will eventually prove compact manifolds may be embedded into Euclidean space by taking a finite cover of open sets, constructing local embedding maps from each open set to Euclidean space, then using a partition of unity to stitch together these maps and create an embedding of the manifold into Euclidean space.

[4.7.7]Theorem(Existence of Partition of Unity)#

Let $\{U_1, \dots, U_n\}$ be a finite open covering for the normal space $X$ . Then $X$ admits a partition of unity dominated by $\{U_k\}_{k = 1}^n$ .

Proof.

We first show that any finite cover $\{U_k\}_{k = 1}^n$ of $X$ admits a finite cover $\{V_k\}_{k = 1}^n$ such that $\bar{V_k} \subseteq U_k$ for each $k \in \{1, \dots, n\}$ . Define

A = U \setminus (U_2 \cup \cdots U_n).

Note that $A$ is a closed subset of $U_1$ , and by normality, there exists an open set $V_1$ such that $A \subseteq V_1 \subseteq \bar{V_1} \subseteq A$ . Thus $\{V_1, U_2, \dots, U_n\}$ is a finite open cover for $U$ , since the missing elements from the incomplete cover $\{U_2, \dots, U_n\}$ are in $A$ , which is a subset of $V_1$ . In general, if

\{V_1, \dots, V_{k - 1}, U_k, U_{k + 1}, \dots, U_n\}

covers $X$ , let

A = X \setminus (V_1 \cup \cdots \cup V_{k - 1} \cup U_{k + 1} \cup \cdots \cup U_{n}).

Apply the same procedure to find $V_k$ , then note $\{V_1, \dots, V_k, U_{k + 1}, \cdots, U_n\}$ covers $X$ . Induction up to the $n$ -th step shows this result is true.

Finally, let $X$ be a normal space together with a finite open cover $\{U_1, \dots, U_n\}$ . By the previous theorem, choose $\{V_k\}_{k = 1}^m$ and $\{W_k\}_{k = 1}^m$ such that $\bar{V_k} \subseteq U_k$ and $\bar{W_k} \subseteq V_k$ for each $k \in \{1, \dots, n\}$ . Note that $\bar{W_k}$ and $X \setminus V_k$ are disjoint, closed sets in a normal space $X$ , so Urysohn's lemma guarantees a continuous map $\psi_k: X \to [0, 1]$ such that $\psi_k(\bar{W_k}) = \{1\}$ and $\psi_k(X \setminus V_k) = \{0\}$ . Thus $\psi_k^{-1}(\bR \setminus \{0\}) \subseteq V_k$ , and so

\supp(\psi_k) \subseteq \bar{V_k} \subseteq U_k.

Define $\psi(x) = \sum_{k = 1}^n \psi_k(x)$ , which is strictly positive (find $k$ such that $x \in U_k$ , then note $\psi_k(x) > 0$ ). Then define

\phi_k(x) = \frac{\psi_k(x)}{\psi(x)},

and note $\{\phi_k\}_{k = 1}^n$ is a partition of unity for $X$ .

[4.7.8]Theorem#

A compact manifold can be embedded into finite-dimensional Euclidean space.

Proof.

Let $X$ be a compact $m$ -manifold. Note that by compactness and by $m$ -manifolds being locally homeomorphic to $\bR^m$ , there exists a finite open cover $\{U_k\}_{k = 1}^n$ comprised of open sets that are homeomorphic to open subsets of $\bR^m$ . Thus, for each $k \in \{1, \dots, n\}$ , there exists a map $g_k: U_k \to \bR^m$ that is an embedding.

Since $X$ is a compact manifold, it is compact Hausdorff and thus normal, so it admits a partition of unity $\{\phi_k\}_{k = 1}^n$ dominated by $\{U_k\}_{k = 1}^n$ . For each $k \in \{1, \dots, n\}$ , define $h_k: X \to \bR^m$ by

h_k(x) = \begin{cases} \phi_k(x) \cdot g_k(x) & x \in U_k, \\ \vb{0}_m & x \in X \setminus \supp(\phi_k). \end{cases}

Note that on the intersection $U_k \cap (X \setminus \supp(\phi_k))$ , we have $\phi_k(x) = 0$ , so both branches agree with $\vb{0}_m$ . Because $U_k$ and $X \setminus \supp(\phi_k)$ are open sets whose union is $X$ , each $h_k$ is well-defined and continuous by the pasting lemma.

Finally, set $N = n + nm$ and define the global map

g: X \to \bR^N \cong \left( \underbrace{\bR \times \cdots \times \bR}_{n\text{ times}} \times \underbrace{\bR^m \times \cdots \times \bR^m}_{n\text{ times}} \right)

g(x) = (\phi_1(x), \dots, \phi_n(x), \, h_1(x), \dots, h_n(x)).

We show that $g$ is an embedding of $X$ into $\bR^N$ . At once we have that $g$ is continuous, since each of its component functions $\phi_k$ and $h_k$ is continuous.

Next, we show that $g$ is injective. Suppose $x, y \in X$ such that $g(x) = g(y)$ . This implies $\phi_k(x) = \phi_k(y)$ and $h_k(x) = h_k(y)$ for every $k \in \{1, \dots, n\}$ . Because $\{\phi_k\}$ is a partition of unity, we have $\sum_{k=1}^n \phi_k(x) = 1$ , so there must exist at least one index $k$ for which $\phi_k(x) > 0$ . Since $\phi_k(y) = \phi_k(x) > 0$ , both points $x$ and $y$ lie in $\supp(\phi_k) \subseteq U_k$ . Evaluating $h_k$ on $U_k$ yields

\phi_k(x) \cdot g_k(x) = h_k(x) = h_k(y) = \phi_k(y) \cdot g_k(y).

Since $\phi_k(x) = \phi_k(y) > 0$ , we may divide by this nonzero scalar to obtain $g_k(x) = g_k(y)$ . Because $g_k$ is an embedding on $U_k$ , it is injective, forcing $x = y$ . Thus, $g$ is an injective continuous map.

Lastly, because $X$ is compact and $\bR^N$ is Hausdorff, the continuous injective map $g: X \to g(X)$ is a closed map, and therefore a homeomorphism onto its image. This completes the proof that $g$ is an embedding into $\bR^N$ .