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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?

EEEB
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  • higly related –  Feb 20 '18 at 17:04
  • If you can't remember which is which, just test whether the property is true for the discrete or indiscrete topology, and this will give you an idea of what you want. In the first case you want the finest since giving $Y$ the indiscrete topology makes $f$ continuous; in the second case you want coarsest since giving $X$ the discrete topology makes it continuous. – MT_ Feb 20 '18 at 17:18

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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".

Caelan Ravel
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