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So let $f:S^{m}\to X$ be a continuous mapping. How can I prove that a) $f$ is homotopic with the constant mapping (i.e. the point I guess) and b) that $f$ can be augmented to a new mapping $f': D^{m+1}\to X$ (i.e. from the unit disk in $m+1$ dimensions to $X$) are equivalent?

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  • $\begingroup$ In general $f$ will not be homotopic to a constant map. Do you have more information about $f$? Also, what is $Y$? Related to $X$ in any way? $\endgroup$ – Alex G. Nov 21 '14 at 14:17
  • $\begingroup$ Sorry I made a typo $Y=X$. As for $f$ I know nothing more. I just think that I have to show that the sentences a) and b) are equivalent. Not to show that each one it true by itself. Does this make it easier? $\endgroup$ – Marion Nov 21 '14 at 14:22
  • $\begingroup$ You should edit your question in that case $\endgroup$ – Alex G. Nov 21 '14 at 14:23
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A hint to get you going: Let $H:S^m \times I \to X$ be a homotopy from $f$ to a constant map. Then $H$ is constant when restricted to $S^m \times \{1\}$, so it factors through the quotient space $S^m\times I / S^m \times \{1\}$ (i.e. it induces a map from this space to $X$). But that quotient space is exactly $D^{m+1}$! Can you see why?

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  • $\begingroup$ Hmm no. I think I am more confused actually :/ Can you elaborate it a bit more please? $\endgroup$ – Marion Nov 21 '14 at 14:36
  • $\begingroup$ The quotient space I described is what is called the "cone" on a space. Try to imagine the cone on $S^1$, and convince yourself that it is the same as $D^2$. Generally, the cone on $S^m$ is homeomorphic to $D^{m+1}$ $\endgroup$ – Alex G. Nov 21 '14 at 14:39
  • $\begingroup$ Ok I am confused. You show that $f$ is homotopic with a constant map because you define a homotopy that is restricted to $S^m \times \{1\}$ which is the srinking of a circle to a point, right? $\endgroup$ – Marion Nov 21 '14 at 14:46
  • $\begingroup$ But the cone of $X$ which is $CX$ is the product space $X \times [0,1]/equivalency class$. Why have you put it to be $X \times \{1\}$? $\endgroup$ – Marion Nov 21 '14 at 14:49
  • $\begingroup$ That notation means that we identify all points in $X\times\{1\}$ with each other, and make no other identifications. $\endgroup$ – Alex G. Nov 21 '14 at 14:50

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