( Homotopy) Y ENR and contractible subset => Y is a retract - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:39:41Z http://mathoverflow.net/feeds/question/119385 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119385/homotopy-y-enr-and-contractible-subset-y-is-a-retract ( Homotopy) Y ENR and contractible subset => Y is a retract Clara 2013-01-20T04:35:39Z 2013-02-09T05:20:16Z <p>I'm trying to solve the following question:</p> <p>Y $\subset R^n$ is a euclidian neighborhood retract. </p> <p>I want to prove that if $Y$ is contractible it is a rectract of $R^n.$</p> http://mathoverflow.net/questions/119385/homotopy-y-enr-and-contractible-subset-y-is-a-retract/119388#119388 Answer by Wlodzimierz Holsztynski for ( Homotopy) Y ENR and contractible subset => Y is a retract Wlodzimierz Holsztynski 2013-01-20T08:12:44Z 2013-02-09T05:20:16Z <p>Let me solve the case of compact $Y$. Thus $Y$ is a contractible ANR. Then Y is an AR.</p> <p><strong>Indeed</strong>, the equivalence&nbsp; "contractible ANR $\Leftrightarrow$ AR"&nbsp; is a well known theorem, certainly known in the past to the founder of the theory of ANRs, Karol Borsuk. To prove this equivalence, consider an arbitrary compact metric space $X$, its closed subset $A$, and an arbitrary continuous map $f : A \rightarrow Y$, where $Y$ is a contractible ANR. Let&nbsp; $H:Y\times[0;1]\rightarrow Y$&nbsp; be a contraction to a point&nbsp; $p\in Y$, meaning that&nbsp; $H(y\ 0)=p$&nbsp; and&nbsp; $H(y\ 1)=y$&nbsp; for every&nbsp; $y\in Y$. Now we get the constant map&nbsp; $g_0:X\rightarrow Y$,&nbsp; such that&nbsp; $\forall_{x\in X}\ g_0(x)=p$,&nbsp; and a homotopy&nbsp; $\Phi : A\times [0;1]\rightarrow Y$ defined by: $$\forall_{a\in A}\forall_{t\in [0;1]}\quad \Phi(a\ t) := H(f(a)\ t)$$ Observe that&nbsp; $\forall_{a\in A}\ \ g_0(a)=\Phi(a\ 0)$. Thus by the Borsuk homotopy extension theorem there exists a homotopy&nbsp; $F:X\times [0;1]\rightarrow Y$&nbsp; such that</p> <ul> <li>$\forall_{x\in X}\quad F(x\ 0)=g_0(x)$ &nbsp; &nbsp; (the value is&nbsp; $p$,&nbsp; of course)</li> <li>$\forall_{a\in A}\forall_{t\in [0;1]}\quad F(a\ t) = \Phi(a\ t)$</li> </ul> <p>Now define&nbsp; $g_1:X\rightarrow Y$&nbsp; by&nbsp; $\forall_{x\in X}\ \ g_1(x) := F(x\ 1)$.&nbsp; Then&nbsp; $g_1$&nbsp; is a continuous extension of&nbsp; $f:A\rightarrow Y$&nbsp; onto $X$. Thus we have proven that $Y$ is an AR.</p> <p>It follows that $Y$ is a retract of the whole $\mathbf R^n$. <strong>Indeed</strong>, space $Y$--being an AR--is a retract of a cube&nbsp; $Q := [-\alpha;\alpha]^n$&nbsp; which contains $Y$, while cube $Q$&nbsp; is a retract of $\mathbf R^n$.</p> <p><strong>REMARK</strong> &nbsp;The last argument was simple, correct and adequate but ad hoc. Here is a more basic (general) argument: let compact $Y\subseteq \mathbf R^n$ be and AR (for metric compact spaces). The Euclidean space $\mathbf R^n$ is a subspace of a metric compact space $C$ (e.g. of&nbsp; $\mathbf R^n\cup{\infty} = \mathbf S^n$),&nbsp; and $Y$ is a retract of $C$, hence $Y$ is a retract of $\mathbf R^n$.</p>