Inducing differentiable structure via continuous maps Consider a map $f: (X,\mathcal{O}_X) \to Y$, with the domain being a topological space with topology $\mathcal{O}_X$ and the codomain merely a set $Y$. We can induce many  topologies on $Y$, however, the most natural one is the finest (or coarsest, I can never remember which is which) topology that "just" makes $f$ continuous: 
$$
\mathcal{O}_Y:=\{B\subset Y \ | \ f^{-1}(B)\in \mathcal{O}_X\}.
$$
With this definition of induced topology on $Y$, $f$ is by definition continuous.
Similarly, for $g : X \to (Y,\mathcal{O}_Y)$, we induce on $X$ a topology that "just" makes $g$ continuous
$$
\mathcal{O}_X=\{g^{-1}(C) \ | \ C\in \mathcal{O}_Y \}.
$$
According to Wikipedia , there is  a good reason why we define topologies with inverse images; inverse images behave well under unions and intersections.

My question is this: can we proceed the same way to induce differentiable structure from a manifold to a topological space? Is there a natural way that we can induce a differentiable structure from a manifold $(M,\mathcal{O}_M, \Phi)$ to $(N,\mathcal{O}_N)$ (topological space) given a continuous $f$ and vice versa given some continuous $g$? Do we do the same thing as in topology, namely define the differentiable structure somehow with inverse images?


That's a neat idea, but I don't think it can work as stated. Here's why:
Think of $f$ as defining an equivalence relation on $M$ where $a\sim b$ whenever $f(a)=f(b)$. Then the coarsest topology on $X$ in which $f$ is continuous will be the quotient topology $M/\sim$, for which $f$ will be the quotient map (which is true whether or not $M$ is a manifold). This new space is not guaranteed to be a manifold. 
For example, if $M=\mathbb R$ and $f(x)=f(y)$ iff $\{x,y\}=\{-1,1\}$, so that the quotient space "looks like" a line looped over itself. This cannot have a locally Euclidean structure at the "crossing point".
