Invariance of domain for manifold 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 ) .
 A: 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. 
