Let $M$ be an $m$-dimensional manifold and $N$ be an $n$-dimensional manifold with boundary. 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.

Intuitively it seems that every continuous function $f:M\rightarrow N$ can be approximated by continuous functions of the form $g:M\rightarrow\operatorname{int}(N)$; where $\operatorname{int}(N)=N- \partial N$, $\partial N$ denoting the boundary of $N$. But is this formally true? I.e., is it true that

$$ \overline{C(M,\operatorname{int}(N))} = C(M,N)? $$

Prototype construction: Let $N=[0,b)$ then any function $f$ can be approximated by: $$ f_n= \min\left(\frac1{n},f\right) , $$ of course, these are continuous but not smooth (since we don't need smoothness this is not an issue). I expect this type of construction can be generalized.


A boundary of a paracompact manifold has a collar neighborhood, i.e. $U\subset N$ that includes $\partial N$ and is homeomorphic to $\partial N\times [0,1)$ via a map $\psi$ that maps $\partial N$ onto $\partial N\times \{0\}$. Therefore, I will be talking about the points in $U$ as if they were in $\partial N\times [0,1)$.

For $n>1$ define a sequence of continuous maps $\varphi_{n}:N\to int(N)$ which is the identity on $N\backslash (\partial N\times [0,\frac{1}{n}))$, and such that $\varphi_{n}(x,t)=(x,\frac{1}{n})$, if $(x,t)\in \partial N \times [0,\frac{1}{n})$. Clearly, $\varphi_{n}$ converges to the identity map in the compact-open topology.

Now, if $f:M\to N$, the sequence $\varphi_n\circ f$ converges to $f$ in the compact open topology, since composition of map is a continuous operation with respect to the compact-open topology.

Note that the fact that $M$ is a manifold is not used, while the fact that $N$ is a manifold is used rather lightly. I wonder under which condition on a connected metric spaces $M$, $N$ and $F\subset N$ the set $C(M,N\backslash F)$ is dense in $C(M,N)$?

Originally (without putting much thought into it) I suggested $F$ to be merely closed nowhere dense and not separating $N$, but @Pietro Majer swiftly refuted that "conjecture".

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    $\begingroup$ As to the last question, consider $N=M=D$, the closed unit disk of $\mathbb{R}^2$, and $F=\{0\}$. Then any map $f:D\to D$ close to the identity, say $\|f-\text{id}\|_{\infty,D} <1$ has degree $1$ wrto $0$, so it can't be in $C(D,D\setminus\{0\})$. $\endgroup$ – Pietro Majer Mar 7 at 8:56

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