I'm trying to solve the following question:
Suppose $Y \subset R^n$ is a Euclidean neighborhood retract. I want to prove that if $Y$ is contractible, then it is a retract of $R^n$.
I'm trying to solve the following question: Suppose $Y \subset R^n$ is a Euclidean neighborhood retract. I want to prove that if $Y$ is contractible, then it is a retract of $R^n$. 


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
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 $\mathbf R^n$. Indeed, space $Y$being an ARis a retract of a cube $Q := [\alpha;\alpha]^n$ which contains $Y$, while cube $Q$ is a retract of $\mathbf R^n$. 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 $\mathbf R^n$. 


Observe that any retract of $\newcommand{\RR}{\mathbb{R}} \RR^n$ is necessarily a closed subspace of $\RR^n$. Assuming this necessary condition, the answer to the question is affirmative. More precisely, if $Y$ is a contractible ANR (absolute neighbourhood retract) and a closed subspace of $\RR^n$, then $Y$ is a retract of $\RR^n$. This is an instance of the following result. Claim: Let $X$ be a metrizable ANR, and let $Y$ be an ANR and a closed subspace of $X$. If the inclusion of $Y$ into $X$ is a homotopy equivalence, then $Y$ is a retract of $X$. This claim is an immediate consequence of the next two propositions. Proposition 1: Let $X$ be a metrizable ANR, and let $Y$ be a closed subspace of $X$. If $Y$ is an ANR, then the inclusion of $Y$ into $X$ is a cofibration. This is stated as proposition A.6.7 in the appendix to Fritsch and Piccinini's book "Cellular structures in topology", page 282. It is also stated and proved in SzeTsen Hu's 1965 book "Theory of retracts": see theorem 3.2 and corollary 3.3 on pages 120 and 121. Alternatively, you may look at the single lemma and its proof in this other answer, which gives proposition 1 for the case of ENRs. Proposition 2: Assume $Y$ is a subspace of $X$ such that the inclusion of $Y$ into $X$ is a cofibration and a homotopy equivalence. Then $Y$ is a strong deformation retract of $X$; in particular, $Y$ is a retract of $X$. This is a standard result in basic homotopy theory. For example, it is stated and proved as corollary 0.20 in chapter 0 of Allen Hatcher's book "Algebraic topology". 

