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Dan Ramras
Dan Ramras
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This answer is a correction to my comment above. In the notesthese notes from Marshall Cohen's course (from Spring 2001), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$.

This answer is a correction to my comment above. In the notes from Marshall Cohen's course (from Spring 2001), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$.

This answer is a correction to my comment above. In these notes from Marshall Cohen's course (from Spring 2001), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$.

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Dan Ramras
Dan Ramras
  • 8.8k
  • 3
  • 47
  • 77

This answer is a correction to my comment above. In the notes from Marshall Cohen's course (from Fall 2000, I believeSpring 2001), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$.

This answer is a correction to my comment above. In the notes from Marshall Cohen's course (from Fall 2000, I believe), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$.

This answer is a correction to my comment above. In the notes from Marshall Cohen's course (from Spring 2001), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$.

Source Link
Dan Ramras
Dan Ramras
  • 8.8k
  • 3
  • 47
  • 77

This answer is a correction to my comment above. In the notes from Marshall Cohen's course (from Fall 2000, I believe), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$.