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acknowledge and discuss shortcoming in answer
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Jeff Strom
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Let $f:S^n\to C$ and let $K$ be its image. If $n>1$ then the Theorem implies $f\simeq *$.

EDIT: Clearly this is over-simple and wrong, as Omar Antolín-Camarena has pointed out. In fact, it looks like $C^N(T)$ would be a compact CW complex, and so there will be surjective continuous functions $S^n \to C^N(T)$ for any $n\geq 1$; then $K = C^N(T)$.

Let $f:S^n\to C$ and let $K$ be its image. If $n>1$ then the Theorem implies $f\simeq *$.

Let $f:S^n\to C$ and let $K$ be its image. If $n>1$ then the Theorem implies $f\simeq *$.

EDIT: Clearly this is over-simple and wrong, as Omar Antolín-Camarena has pointed out. In fact, it looks like $C^N(T)$ would be a compact CW complex, and so there will be surjective continuous functions $S^n \to C^N(T)$ for any $n\geq 1$; then $K = C^N(T)$.

Source Link
Jeff Strom
  • 12.5k
  • 3
  • 48
  • 76

Let $f:S^n\to C$ and let $K$ be its image. If $n>1$ then the Theorem implies $f\simeq *$.