Approximation and homotopy

Question

In Differential Topology the usual strategy to discuss homotopy classes by differential means is approximation, because a manifold $M$ is a euclidean neigborhood retract (ENR) and any good approximation of a map $f:X\to M$ is homotopic to $f$. Now, consider the following situation where this does not work: a pair of manifolds $Z\subset Y$, $Z$ not closed in $Y$; more precisely, suppose that $M=Y\setminus Z$ is not locally compact, hence not a ENR and we cannot guarantee approximation implies homotopy (as far as I know). The question is to exhibit an explicit counterexample: a continuous map $f:X\to M$ which is not homotopic to its smooth close approximations.

To give some concrete example, suppose $Y=\mathbb R^4$, $\dim(Z)=1$ to have enough room for manipulations. We recall $Z$ is not closed in $\mathbb R^4$, even more, $M=\mathbb R^4\setminus Z$ is not locally compact. The typical transversality attempt to show $M=Y\setminus Z$ is simply connected would be as follows:

(1) Every continuous loop $f:\mathbb S^1\to M$ can be arbitrarily approximated by a smooth loop $g:\mathbb S^1\to\mathbb R^4$ transversal to $Z$ (note the codomain of $g$ is not $M$), and since cod$(Z)\ge2$ transversality means $g$ does not meet $Z$ and so actually $g:\mathbb S^1\to M$.

(2) Now, $g$ is nulhomotopic by the smooth homotopy $H_t=(1-t)g+ta$ for any fixed $a\notin Z$. Again, we can approximate $H$ by a new smooth homotopy $\widetilde H$ transversal to $Z$, and since $H_0=f$ and $H_1\equiv a$ are already transversal to $Z$ (they do not meet $Z$!), we can preserve $\widetilde{H_0}=H_0$ and $\widetilde{H_1}=H_1$. Finally since cod$(Z)\ge3$, transversality means $\widetilde H$ does not meet $Z$, and so $\widetilde H$ is a homotopy in $M$, and $g$ is nullhomotopic in $M$.

Summing up, $f$ is arbitrarily close to $g$ and $g$ is nulhomotopic, but we cannot conclude $f$ is nulhomotopic, because approximation does not imply homotopy. Hence a more concrete version of my question above: there is a smooth curve $Z$ not closed in $\mathbb R^4$ whose complement is not simply connected?

Answer 1

As requested, I will answer your last ("concrete") question. In fact, I will prove more:

Theorem. Let $h: M\to N$ be a smooth map of an $m$-manifold to an $n$-manifold, where $n-m=k\ge 2$. Suppose that $N$ is $k-2$-connected. Then $C:=N\setminus h(M)$ is also $k-2$-connected.

Proof. I will equip $N$ with a background metric. Suppose that $f: \partial B^{k-1}\to C$ is a (continuous) map. I will prove existence of a (continuous) extension of $f$ to $B=B^{k-1}$. For a subset $N'\subset N$ let $C^0(B, \partial B; N')$ denote the space of continuous extensions of $f$ to maps $B\to N'$, equipped with topology of uniform convergence. Since $N$ is $k-2$-connected, the space $C^0(B, \partial B; N)$ is nonempty.

I will prove that $C^0(B, \partial B; C)$ is nonempty by proving that it is a $G_\delta$-subset.

Let $(M_i)_{i\in \mathbb N}$ denote an exhaustion of $M$ by a compact codimension zero submanifolds with (smooth) boundary. Set $N_i:= N\setminus h(M_i)$; this is an open subset of $N$. By smooth approximation and transversality, every continuous extension $f$ to a map $B\to N$ is a $C^0$-uniform limit of maps $B^{k-1}\to N_i$. Thus, the subset $C^0(B, \partial B; N_i)$ is open and dense in $C^0(B, \partial B; N)$. But $$ C^0(B, \partial B; C)= \bigcap_{i\ge 1} C^0(B, \partial B; N_i). $$ Hence, $C^0(B, \partial B; C)$ is $G_\delta$-subset, therefore, is nonempty by the Baire Theorem. qed

Note: This proof is inspired by the argument in the 2nd answer here.