Smooth manifold's definition and initial topology

Question

Suppose $M$ is a smooth manifold. Then it has a topology on it (I'm using Lee's notion of smooth manifold). Well, suppose that $\phi_\alpha: U_\alpha \to \mathbb{R}^n$ is a coordinate chart in the maximal atlas. Take on $M$ the coarsest topology making any $\phi_\alpha$ continuous. Is such a topology (i.e. the initial topology induced by $\phi_\alpha$) the same as the topology induced by the smooth structure?

Answer 1

Suppose $M$ is a $0$-dimensional smooth manifold with maximal atlas $\{\varphi_\alpha\}$ and $\mu = \tau(\{\varphi_\alpha^{-1}(V) : V\subseteq \mathbb{R}^0\text{ is open}\})$ is the standard topology on $M$ (here $\tau(\mathcal{A})$ denotes topology generated by $\mathcal{A}$), so that $\mu$ is the discrete topology, and consider indiscrete topology $\tau = \{\emptyset, M\}$ on $M$. Then $U_\alpha = \{x_\alpha\}$ are one-point sets, and so $\varphi_\alpha:(U_\alpha, \tau\restriction_{U_\alpha}) = \{x_\alpha\}\to \mathbb{R}^0$ are trivially continuous, but unless $|M|\leq 1$, the indiscrete topology does not agree with the discrete topology on $M$. This shows that the smallest topology on $M$ which makes the charts $\varphi_\alpha$ continuous, which is $\tau$, does not coincide with $\mu$.

Its interesting to ask what happens when $n \geq 1$.

---

Note however that if $\tau$ is the smallest topology for which $\varphi_\alpha$ are continuous and $U_\alpha$ are open, then $\tau \subseteq \mu$, but also since given open $V\subseteq \mathbb{R}^n$, the set $\varphi_\alpha^{-1}(V)\subseteq U_\alpha$ is open in $(U_\alpha, \tau\restriction_{U_\alpha})$ and $U_\alpha$ is open in $(M, \tau)$, it follows that $\varphi_\alpha(V)^{-1}\in \tau$. Since sets of the form $\varphi_\alpha^{-1}(V)$ generate the topology $\mu$, we have $\mu\subseteq \tau$, so $\mu = \tau$.