2 Another, this time more stylish final argument (earlier I was lazy :-).

Let me solve the case of compact $Y$. Thus $Y$ is a contractible ANR. Then Y is an AR.

Indeed, the equivalence  "contractible ANR $\Leftrightarrow$ AR"  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  $H:Y\times[0;1]\rightarrow Y$  be a contraction to a point  $p\in Y$, meaning that  $H(y\ 0)=p$  and  $H(y\ 1)=y$  for every  $y\in Y$. Now we get the constant map  $g_0:X\rightarrow Y$,  such that  $\forall_{x\in X}\ g_0(x)=p$,  and a homotopy  $\Phi : A\times [0;1]\rightarrow Y$ defined by: $$\forall_{a\in A}\forall_{t\in [0;1]}\quad \Phi(a\ t) := H(f(a)\ t)$$ Observe that  $\forall_{a\in A}\ \ g_0(a)=\Phi(a\ 0)$. Thus by the Borsuk homotopy extension theorem there exists a homotopy  $F:X\times [0;1]\rightarrow Y$  such that

• $\forall_{x\in X}\quad F(x\ 0)=g_0(x)$     (the value is  $p$,  of course)
• $\forall_{a\in A}\forall_{t\in [0;1]}\quad F(a\ t) = \Phi(a\ t)$

Now define  $g_1:X\rightarrow Y$  by  $\forall_{x\in X}\ \ g_1(x) := F(x\ 1)$.  Then  $g_1$  is a continuous extension of  $f:A\rightarrow Y$  onto $X$. Thus we have proven that $Y$ is an AR.

It follows that $Y$ is a retract of the whole $R^n$. \mathbf R^n$. Indeed, space$Y$--being an AR--is a retract of a cube$Q := [-\alpha;\alpha]^n$which contains$Y$, while cube$Q$is a retract of$R^n$.\mathbf R^n$.

REMARK  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  $\mathbf R^n\cup{\infty} = \mathbf S^n$),  and $Y$ is a retract of $C$, hence $Y$ is a retract of $\mathbf R^n$.

1

Let me solve the case of compact $Y$. Thus $Y$ is a contractible ANR. Then Y is an AR.

Indeed, the equivalence  "contractible ANR $\Leftrightarrow$ AR"  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  $H:Y\times[0;1]\rightarrow Y$  be a contraction to a point  $p\in Y$, meaning that  $H(y\ 0)=p$  and  $H(y\ 1)=y$  for every  $y\in Y$. Now we get the constant map  $g_0:X\rightarrow Y$,  such that  $\forall_{x\in X}\ g_0(x)=p$,  and a homotopy  $\Phi : A\times [0;1]\rightarrow Y$ defined by: $$\forall_{a\in A}\forall_{t\in [0;1]}\quad \Phi(a\ t) := H(f(a)\ t)$$ Observe that  $\forall_{a\in A}\ \ g_0(a)=\Phi(a\ 0)$. Thus by the Borsuk homotopy extension theorem there exists a homotopy  $F:X\times [0;1]\rightarrow Y$  such that

• $\forall_{x\in X}\quad F(x\ 0)=g_0(x)$     (the value is  $p$,  of course)
• $\forall_{a\in A}\forall_{t\in [0;1]}\quad F(a\ t) = \Phi(a\ t)$

Now define  $g_1:X\rightarrow Y$  by  $\forall_{x\in X}\ \ g_1(x) := F(x\ 1)$.  Then  $g_1$  is a continuous extension of  $f:A\rightarrow Y$  onto $X$. Thus we have proven that $Y$ is an AR.

It follows that $Y$ is a retract of the whole $R^n$. Indeed, space $Y$--being an AR--is a retract of a cube  $Q := [-\alpha;\alpha]^n$  which contains $Y$, while cube $Q$  is a retract of $R^n$.