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Let $M$ be an $m$-dimensional manifold and $N$ be an $n$-dimensional manifold. Suppose also that the topology on $N$ can be described by a metric. Thus, the set $C(M,N)$ can be endowed with the topology of [uniform convergence on compacta][2].

Let $N'\subseteq N$ be a dense subset which is homeomorphic to $\mathbb{R}^n$. In this post's answer's comment it was remarked that $C(D,D-\{0\})$ is not dense in $C(D,D)$ where $D$ is the unit disc.

In general, when is $C(M,N')$ not dense in $C(M,N)$?

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    $\begingroup$ Actually, even the case of $N=S^1$, $N'=\mathbb{R}$ is an example of non-density. Note that if $N'$ is homeomorphic to $\mathbb{R}^n$, every map into $N'$ is null-homotopic. Null-homotopic maps are closed in $C(M,N)$, and so $C(M,N')$ is NOT dense as long as there is a non-null homotopic map from $M$ into $N$. So, if $N$ is not contractible, the identity map cannot be approximated. I am however not that well-versed in algebraic topology to prove that if $N$ is null-homotopic, $C(M,N')$ is dense. $\endgroup$ – erz Apr 2 at 0:29
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    $\begingroup$ Would you know a reference where I can find the fact that null-homotopic maps form a closed subspace? $\endgroup$ – MrMMS Apr 2 at 6:43
  • $\begingroup$ Yeah, I thought that it is obvious when I was writing that comment, but now I cannot find neither a proof, nor a reference. I guess, time to ask yet another question. $\endgroup$ – erz Apr 2 at 22:35
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1) About Erz' answer, and Annie's second question. It is a general property that for two differentiable manifolds $M$, $N$, such that $M$ is compact, and given a continuous map $f:M\to N$, any continuous map $g:M\to N$ close enough to $f$ is homotopic to $f$. This can be proved by many different ways, depending on taste. (One can assume that $N$ is also compact, since we are only interested in a small neighborhood of the compact $f(M)$ in $N$):

a) Put a Riemannian metric on $N$, let $i>0$ be its injectivity radius: any two points $x,y\in N$ whose distance is less than $i$ are linked by a unique shortest geodesic, and this geodesic depends continuously on the pair $(x,y)$. Then, assuming that $d(f(x),g(x))<i$ for every $x\in M$, the geodesics from $f(x)$ to $g(x)$ provide the desired homotopy;

b) Embed $N$ in $R^n$ for $n$ large enough (Whitney); then $N$ has a tubular neighborhood $T$ in $R^n$ together with a continuous retraction $p:T\to N$; if $g$ is close enough to $f$, then the segments $[f(x),g(x)]$ in $R^n$ are contained in $T$ and their projection under $p$ provide the homotopy;

c) Triangulate $N$ and use the straight lines in the simplices of the triangulation instead of the geodesics;

d) Use a good cover of $N$ by open subsets $U(i)$ such that every intersection $U(i_1)\cap\dots\cap U(i_k)$ which is nonempty is contractible...

2) Erz' homotopical argument is excellent; but even if $N$ is contractible, $C(M,N')$ is in general not dense in $C(M,N)$. We should think to the complement $X$ of $N'$ in $N$. For example, let $M$ be the compact interval $[-1,+1]$, $N$ be the complex plane $\mathbb{C}$ and $X$ be the nonpositive real halfline $\mathbb{R}_-$. Clearly, $\mathbb{C}\backslash X$ is diffeomorphic with the plane (being open and starred with respect to $1$.) Then, the embedding $[-1,+1]\to\mathbb{C}$: $t\mapsto-1+it$ can certainly not be approximated by continuous maps valued in $\mathbb{C}\backslash X$ (by the elementary "intermediate value theorem").

More generally, note that by Weierstrass' approximation theorem, Annie's question is equivalent to: when is $C^\infty(M,N')$ not $C^0$-dense in $C^\infty(M,N)$? The advantage of this differentiable setting is that thanks to Thom's transversality theorem, one can answer in all cases where $X$ is regular enough (technically, a finite or denumerable union of smooth submanifolds $S_i\subset N$; e.g. a polytope):

a) If one at least of the $S_i$'s is of codimension $\le m$ in $N$, then there will be a smooth map $f:M \to N$ which intersects $S_i$ transversely in at least one point; and $f$ cannot be in the $C^0$-closure of $C(M,N')$.

b) If on the contrary each $S_i$ is of codimension $>m$ in $N$, then by Thom's transversality theorem, $C^\infty(M,N')$ is $C^\infty$-dense in $C^\infty(M,N)$, and in particular $C^0$-dense. (As to Erz' second question: since $\dim(M)+\dim(S_i)<\dim(N)$, saying that $f:M\to N$ is transverse to $S_i$ amounts to say that $f(M)$ does not meet $S_i$).

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    $\begingroup$ I took liberty of latexifying your answer. I hope you don't mind. Could you please elaborate on your last point? I looked up Hirsh, and he says that the transversal maps are dense. But why are "avoiding" maps dense in the set of transversal ones? $\endgroup$ – erz Apr 3 at 20:58
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    $\begingroup$ Also, what if we apply this to $M=N$? $X$ then cannot contain a submanifold of any co-dimension, and so should be empty? And finally, is there a way to have similar reasoning without smoothness? $\endgroup$ – erz Apr 3 at 20:59
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    $\begingroup$ I also have a question/ confusion: If $N=[0,1]^n,N'=(0,1)^n, X=\partial (0,1)^n$, and $M=\mathbb{R}^m$; with $m\geq 1$. Then unless $n\leq 1$, $C(M,N')$ is not dense in $C(M,N)$? Doesn't that contradict @erz 's answer to this post: mathoverflow.net/questions/354304/… ? I'm assuming that the argument somehow breakdown when looking at (smooth) manifolds with boundaries? $\endgroup$ – MrMMS Apr 6 at 14:21
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    $\begingroup$ @AnnieTheKatsu From my understanding the argument goes like this: if $X$ contains a submanifold and a $f$ is transversal to this submanifold, then locally, you have the inclusion of $\mathbb{R}^m$ into $\mathbb{R}^m\oplus \mathbb{R}^k$, where $X$ contains $\mathbb{R}^k$. Surely, you cannot approximate such an inclusion avoiding $X$. However, this only works for submanifolds. The boundary of a manifold with boundary is not a submanifold in this sense. $\endgroup$ – erz Apr 6 at 16:58
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    $\begingroup$ Of course, the transversality theorem does not hold when the submanifold is contained in the boundary. As for the other question, by "Weierstrass" I mean: every continuous map $f:M\to N$ between two manifolds can be approximated by a smooth one. This is proved by embedding $N$ into $R^d$ (Whitney), approximating $f$ by a smooth map $g:M\to R^d$ by the classical Weierstrass theorem, and reprojecting $g$ into $M$ by a local projection. $\endgroup$ – Gael Meigniez Apr 7 at 15:21

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