I am trying to prove the following theorem:

Theorem. Every compact manifold is homeomorphic to a subset of some Euclidean space.

The manifolds I'm considering are the most general (without any smooth, PL or any structures). Here's a formal definition:

$n$-Manifolds. An $n$-manifold $\mathcal M$ is a second-countable, Hausdorff topological space such that every point $p\in\mathcal M$ has a neighborhood $U\subseteq\mathcal M$ that is homeomorphic to $\mathbb R^n$.

Here's my (incomplete) attempt at a proof:\

Proof. Suppose $\mathcal M$ is a compact $n$-manifold. By compactness, $\mathcal M$ has a finite open cover, say, $U_1,\dots,U_k$. By definition, we are guaranteed that each of these open subsets is homeomorphic to $\mathbb R^n$. That is, we can choose homeomorphisms $\phi_i:U_i\to\mathbb R^n$ for each $i$. We recall that since $\mathcal M$ is compact, it is also paracompact and hence admits a partition of unity subordinate to this open cover, say, $(\psi_i)_i$. Then I'm lost...

My idea was to use the functions $\phi_i$'s and $\psi_i$'s to construct a homeomorphism $\Phi:\mathcal M\to\mathbb R^j$ for some $j$ using the Pasting Lemma (also known as Gluing Lemma). But I don't know how to use that lemma to construct $\Phi$ and show that is it a homeomorphism. I have already spent over two days in trying to figure this out. Any help would be appreciated. TIA.

  • $\begingroup$ @Jakobian : I'm aware of the famous Whitney's Embedding Theorem. But currently, I am looking for a proof using the above mentioned facts (homeomorphisms and partitions of unity and the pasting lemma) only. $\endgroup$ Jun 21, 2023 at 18:51
  • $\begingroup$ What you are trying to prove, for smooth manifolds is known as Whitney embedding theorem and it is farm from trivial. For compact metric spaces, these can also embedded in Euclidean space due to covering dimension argument, see Hurewitz-Wallman's "Dimension Theory". Again, nontrivial. $\endgroup$
    – freakish
    Jun 21, 2023 at 18:55
  • $\begingroup$ Compact manifolds*. Not all compact metric spaces are finite-dimensional $\endgroup$
    – Jakobian
    Jun 21, 2023 at 19:10
  • $\begingroup$ Yes, I meant compact manifolds. $\endgroup$
    – freakish
    Jun 21, 2023 at 19:28
  • 1
    $\begingroup$ The proofs for explicit bounds on $N$ for which an $n$-dimensional manifold can be embedded into $\mathbb{R}^N$ are hard to prove (e.g. Whitney embedding theorem), but this one is more straightforward (!). $\endgroup$
    – Jakobian
    Jun 21, 2023 at 19:33

1 Answer 1


Take open finite open cover of $X$ by sets homeomorphic to $\mathbb{R}^n$, say $U_1, ..., U_k$. For each $i$, let $f_i:U_i\to \mathbb{R}^n$ be a homeomorphism. By considering the one-point compactification $\mathbb{R}^n\cup\{\infty\}\cong S^n$ of $\mathbb{R}^n$, you can extend $f_i$ to a continuous map $f_i:X\to \mathbb{R}^n\cup\{\infty\}$ where $f_i$ sends $X\setminus U_i$ to $\infty$. Verify this.

The map $f(x) = (f_1(x), ..., f_k(x))$ is then an embedding, since it's a closed injective continuous map. As mentioned above, the one-point compactification of $\mathbb{R}^n$ naturally embedds into $\mathbb{R}^{n+1}$, so this gives an embedding of $X$ into $\mathbb{R}^N$ for some $N\in\mathbb{N}$.

  • $\begingroup$ Thanks for this simple but elegant proof without even invoking the existence of the partitions of unity. This is exactly the kind of proof I was trying to find. By the way, did you mean "$\mathbb{R}^n\cup\{\infty\}\cong S^{n}$" instead of "$\mathbb{R}^n\cup\{\infty\}\cong S^{n+1}$" ? Because it is $\mathbb S^n$ which naturally embeds into $\mathbb R^{n+1}$. $\endgroup$ Jun 21, 2023 at 20:24
  • $\begingroup$ Oh shoot, sorry. Yeah that's what I meant $\endgroup$
    – Jakobian
    Jun 21, 2023 at 20:26
  • $\begingroup$ I just came back with a slight confusion. Can you elaborate on how to show that $f(x)$ defined above is injective? I know why it's closed and continuous but don't know how to check injectivity in this case. $\endgroup$ Jun 21, 2023 at 23:51
  • 1
    $\begingroup$ If $f(x) = f(y)$ and $x\in U_i$, then $f_i(x) = f_i(y)\in\mathbb{R}^n$ so $y\in U_i$. But recall that $f_i$ is a homeomorphism from $U_i$ to $\mathbb{R}^n$, so $x = y$. $\endgroup$
    – Jakobian
    Jun 22, 2023 at 1:13

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