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Let $M$ be a 0-dimensional smooth manifold and $N$ a smooth manifold with or without boundary, is any map $f:M\to N$ smooth?

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Yes. A map $f:M\to N$ is smooth if and only if for all $p\in M$ there are smooth charts $(U,\varphi)$ containing $p$ and $(V,\psi)$ containing $f(p)$ such that $f(U)\subseteq V$ and the composition $\psi\circ f\circ\varphi^{-1}:\varphi(U)\to\psi(V)$ is smooth. In case $M$ is $0$-dimensional and $N$ is $n$-dimensional, we can take $U=\{p\}$, $\varphi:\{p\}\to \{0\}=\mathbb{R}^0:p\mapsto 0$, and any chart $(V,\psi)$ containing $f(p)$. This satisfies the requirement since the map $\psi\circ f\circ\varphi^{-1}:\{0\}=\mathbb{R}^0\to\psi(V)\subseteq\mathbb{R}^n$ is constant and hence smooth.

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This is basically asking if every point of $N$ has a smooth coordinate neighborhood, which is true since $N$ is smooth.

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  • $\begingroup$ I would agree that $f$ must be continuous, since any map out of a discrete space is continuous. Is it conventional to consider maps out of discrete spaces smooth as well? $\endgroup$ – Charles Hudgins Jun 24 at 11:05
  • $\begingroup$ Only if the target space is smooth. Consider the map $f$ that maps your point to a non-smooth point of $N$. $\endgroup$ – Ruben Jun 24 at 11:14

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