# Is any map $f:M\to N$ smooth?

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?

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.
This is basically asking if every point of $$N$$ has a smooth coordinate neighborhood, which is true since $$N$$ is smooth.
• 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? – Charles Hudgins Jun 24 at 11:05
• Only if the target space is smooth. Consider the map $f$ that maps your point to a non-smooth point of $N$. – Ruben Jun 24 at 11:14