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If a space X has with contractible universal cover then any map $S^n> \rightarrow X$ can be extended to $D^{n+1} \rightarrow X$

$D^{n+1}$ is the ball with dimension $n+1$

How should I approach this proof? Ihave drawn a diagram with the spaces, the maps and also the induced maps for the fundamental groups. I see that it should be true, but do not know how to start it.

southernKid33
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  • Is $D^{n+1}$ the ball of dimension $n+1$? – Fimpellizzeri Feb 09 '20 at 22:25
  • Yes, the closed ball. So there is an inclusion map $i:S^n \rightarrow D^{n+1}$ – southernKid33 Feb 09 '20 at 22:27
  • We only need to know that $X$ has some contractible covering space. Note, however, that a contractible covering space is simply connected and thus is a universal covering space. For technical reasons we also need to require that $X$ is connected and locally path-connected. Just apply the lifting theorem https://math.stackexchange.com/q/2180646. – Paul Frost Feb 12 '20 at 15:42

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Let $\hat X$ be the universal cover of $X$ and $p:\hat X\rightarrow X$ the covering map, there exists a family of maps $h_t:\hat X\rightarrow \hat X$ such that $h_0=Id_X, h_1$ is constant. Let $f:S^n\rightarrow X$, we can lift it to a map $\hat f:S^n\rightarrow \hat X$ since $p$ is a Serre fibration $g:D^{n+1}\rightarrow \hat X$ defined by $\hat g(x)=h_{\|x\|}(\hat f({x\over\|x\|}))$, if $x\neq 0, \hat g(x)=x_0$ where $x_0$ is the image of $h_1$. and $g=p\circ\hat g$.

Tsemo Aristide
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  • Where are we using the universal cover? To get the family of maps $h_t$? If you can explain it a little bit, I would appreciate it – southernKid33 Feb 09 '20 at 22:41
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    You do not explicitly need $h_t$. Since $\hat X$ is contractible, the lift $\hat f$ is null-homotopic and thus has a continuous extension $g : D^{n+1} \to \hat X$. – Paul Frost Feb 12 '20 at 15:46