A retract in normal space Let $X$ be a normal space and $K$ be a closed subset of $X$ homeomorphic to $\mathbb R$. Can I always find a retract of $X$ onto $K$?
From some examples, such as $X=\mathbb R^n$ and $K=\mathbb R$, I guess it is true, but I can't find the way to prove it rigorously.

Fix a homeomorphism $h:K\to\mathbb R$.  Let $\overline{\mathbb R}$ denote the extended real line, $\mathbb R\cup\{\infty,-\infty\}$. Since $\overline{\mathbb R}$ is homeomorphic to the interval $[0,1]$ and since $K$ is closed in the normal space $X$, the Tietze extension theorem lets you extend $h$ to a continuous map $\bar h:X\to\overline{\mathbb R}$.  Now let $L=\bar h^{-1}(\{\infty,-\infty\})$ and observe that $L$ is a closed set disjoint from $K$. By normality, find a closed set $M$ with $K\cap M=\varnothing$ and $L\subseteq\text{Int}(M)$, and then use Urysohn's lemma to get a continuous $f:X\to[0,1]$ that is identically 1 on $K$ and identically 0 on $M$. I claim that the function $g:X\to\mathbb R$ defined to be  product $f\cdot\bar h$ on $X-L$ and 0 on $\text{Int}(M)$ is well-defined; the point is that, on the overlap of the two open sets $X-L$ and $\text{Int}(M)$, $f$ is identically zero and $\bar h$ has values in $\mathbb R$, so their product is identically zero. It follows that $g$ is continuous on $X$, because its restrictions to the open sets $X-L$ and $\text{Int}(M)$ are obviously continuous. Also note that $g$ maps to $\mathbb R$, not $\overline{\mathbb R}$, because $L$ is included in the interior of $M$ where $g$ is zero.  Finally, since $f$ is identically 1 on $K$, $g$ coincides with $\bar h$ and thus with $h$ on $K$.  Therefore, the composition $h^{-1}\circ g$ is a retraction of $X$ onto $K$.
