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This is corollary $2B.4$ in Hatcher's Algebraic Topology , state as following :

If $M$ is a compact $n-$ manifold and $N$ is a connected $n-$ manifold then an embedding $h : M \to N$ must be surjective , hence a homeomorphism .

The first , $h(M)$ must be close since $h(M)$ is compact and $N$ is Hausdorff . But I don't know why $h(M)$ is open just from invariance of domain theorem ? If we have $h(M)$ is also open then by connectedness we have $h(M)=N$ hence a homeomorphism .

To be more specific , I'm trying to prove that any continuous map $f : S^{n} \to \mathbb{R^{n}}$ can not be injective ( don't use Borsuk-Ulam theorem ) .

Gankedbymom
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1 Answers1

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Consider and open cover $\mathcal{U}$ of $M$ by coordinate neighbourhood such that the image of each open set in $\mathcal{U}$ under $h$ lies inside a coordinate neighbourhood of $N$ (start with a point in $h(M)$ take a coordinate neighbourhood, pull it back by $h$ and then take an even smaller coordinate neighbourhood inside this and do it for all the points). Let $U\in\mathcal{U}$ with chart $\phi :U\rightarrow R^n.$ Also let $h(U)\subset V$ and $\psi :V\rightarrow R^n$ be the chart in $N$. Since $h$ is an embedding, by invariance of domain, image of $h(U)$ is open (consider the map $\psi h\phi^{-1}:\phi(U)\rightarrow R^n$. By invariance of domain the image (say W) open. As $\psi$ is a homemorphism $\psi^{-1}(W)$ is also open which is equal to $h(U)$). As $h(M)$ is the union of those open sets, $h(M)$ is also open.

Explanation: $h$ is an embedding implies $h$ is homeomorphic onto the image. Therefore if $U\in\mathcal{U}$ then $h(U)$ is open in $h(M)$. Hence $h(U)=h(M)\cap V$ for some open set $V$ in $N$. Cover $V$ by coordinate neighbourhoods $V_{\lambda}$ with chart $\psi_\lambda$. Clearly $V=\bigcup V_\lambda$ hence $h(U)=\bigcup (h(M)\cap V_\lambda)$. Now by invariance of domain the image of $\psi_\lambda h\phi^{-1} :\phi(U)\rightarrow R^n$ is open. This image is equal to $\psi_\lambda(V_\lambda\cup h(M))$. Now $\psi_\lambda$ is a homeomorphism from $N$ to $R^n$ hence $(V_\lambda\cup h(M))$ is open in $N$. $h(U)$, being the union of these open sets, is also open.

tessellation
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  • The problem is why under $h$ each image is also open and I think manifold musn't a subset of $\mathbb{R^{n}}$ which stated by invariance of domain . – Gankedbymom Apr 11 '17 at 18:29
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    This was exactly my point. Manifold is not a subset of $R^n$ but locally manifolds are open subset of $R^n$. That is why I have constructed the coordinate neighbourhood (on bot the manifolds) and the composed it with the chart and $h$ to get a map from open set of $R^n$ to $R^n$ and then applied invariance of domain. – tessellation Apr 11 '17 at 18:32
  • Locally manifold ? You mean compact manifold ? – Gankedbymom Apr 11 '17 at 18:34
  • I have edited the answer. – tessellation Apr 11 '17 at 18:41
  • So the condition countable basis haven't used here ? – Gankedbymom Apr 11 '17 at 18:58
  • No. You don't need countable basis here. But $M$ is compact so automatically have countable basis. – tessellation Apr 11 '17 at 19:11
  • I doubt about the step start with a point in $h(M)$ take a coordinate nbh $(V,\phi)$ ( $V$ maybe not in $h(M)$ ) , restric it to a open set $V'$ in relative topology of $h(M)$ we still obtain a coordinate bhh . Now put $U = h^{-1}(V')$ with chart $\phi . h$ ? Is this true ? It seem that $M$ isn't neccessarily a manifold . – Gankedbymom Apr 12 '17 at 08:04
  • I have added an explanation at the end of the answer. Hope this will clarify your confusion. – tessellation Apr 12 '17 at 08:26
  • :) You have some mistakes at post ( just latex ) . But thank you so much – Gankedbymom Apr 12 '17 at 08:37
  • Could you prove that if $h : U \to M$ is an injection from an open subspace $U$ of manifold $M$ , then $h(U)$ is also open ? I think the proof is same as your post . – Gankedbymom Apr 12 '17 at 11:40
  • If the dimensions are same then yes. – tessellation Apr 12 '17 at 12:22
  • The last sentence , we see $\psi_{\lambda}(V_{\lambda} \cap h(M))$ is open in $R^{n}$ but $\psi_{\lambda}$ just a homeomorphism from $V_{\lambda}$ to $R^{n}$ which means $V_{\lambda} \cap h(M)$ is open in $V_{\lambda}$ . From step cover $V$ , $\forall x \in V , (G_{x},\psi_{x})$ restric to $(K_{x},\psi)$ ( $K_{x}$'s open in $N$) , take $T_{x}=K_{x}\cap V$ is open in $V,N=> T_{x}\cap h(M)$ is open in $T_{x}$ which also means open in $N$ because $T_{x}$ is open in $N$ – Gankedbymom Apr 12 '17 at 13:20
  • $V_\lambda$ is open in $N$. Therefore any open subset of $V_\lambda$ is also open in $N$. This is the definition of subspace topology. – tessellation Apr 12 '17 at 13:39
  • Ok , I just repeat you dont write $V_{\lambda}$ is open in $N$ , the choice isn't natural ? For the question $f : S^{n} \to \mathbb{R^{n}}$ can't be one-to-one . Could you prove it ? Thank you – Gankedbymom Apr 12 '17 at 13:42
  • "Cover $V$ by coordinate neighbourhoods $V_{\lambda}$". By definition a coordinate neighbourhood is open. – tessellation Apr 12 '17 at 15:25
  • :D Ok , sorry because I never see the definition of coordinate neighborhood before ? So as my build each point on manifold always have a coordinate neighborhood ? It that true ? – Gankedbymom Apr 13 '17 at 12:17
  • Each point of a manifold has a coordinate neighbourhood around it. This is in the definition of a manifold along with conditions about transition functions. – tessellation Apr 14 '17 at 07:33