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Lately I’ve been having fun learning about smooth spaces.
Conceptually, I think of it kind of like this…
Given a set $X$, a diffeology $D(X)$ on $X$ tells you which maps into $X$ are smooth (or not smooth).
This is borrowed from the way I used to explain what a topology was to my colleagues (who were top researchers in their field of study but never had the need to know the formal definition of a topology):
Given a set $X$, a topology $T(X)$ on $X$ tells you which subsets of $X$ are open (or closed).
The two statements above are “similar”, but not quite perfect analogies (as far as I can tell).
Is it possible to define a topology on a space in terms of maps into that space? For example,
Given a set $X$, a topology $\mathcal{T}(X)$ on $X$ tells you which maps into $X$ are continuous (or not continuous).
For example, if $f\in\mathcal{T}(X)$ and $g:X\to Y$, we’d say $g$ is continuous if $g\circ f\in\mathcal{T}(Y)$.
Follow up question…
Just as a topology can be defined in terms of open OR closed sets. Could you define a smooth space in terms of maps into $X$ that are NOT smooth? Kind of like the complementary definition of topology in terms of closed sets?
PS: It seems like you can play this game all night long. For example, given a set $X$, a purpleology $P(X)$ tells you which maps into $X$ are purple. If $f\in P(X)$ and $g:X\to Y$, we say $g$ is purple if $g\circ f\in P(Y)$ :)
Could you define a smooth space in terms of maps into X that are NOT smooth?
There is one rule in this game here:
your assignment of sets of allowed maps to a given test space must also respect the morphisms between the test spaces:
if $U$ and $V$ are two test spaces and $X$ is your diffeological space or some other sheaf, and $X(U)$ is the set of allowed ways of mapping $U$ into $X$ and $X(V)$ correspondingly the set of ways of mapping $V$ into $X$, then for every homomorphism of test spaces $\phi : U \to V$ the map that sends $X(V)$ to $X(U)$ by precomposing plots with $\phi$ must
exist (! :-)
respect composition.
Your example of non-smooth maps will violate 1. At least if I understand correctly what you mean:
you could consider some manifold $X$ and declare that $X(U)$ is the set of maps of set $U \to X$ that are not smooth with respect to the smooth structure on $U$ and $X$.
But consider the non-smooth function $H : \mathbb{R} \to \mathbb{R}$ that is constant 0 on the negative ray, constant 1 on the positive ray, and jumps at 0.
Then let $\phi : \mathbb{R} \to \mathbb{R}$ be the map $\exp(-)+1$. Then the composite $\mathbb{R} \stackrel{\phi}{\to} \mathbb{R} \stackrel{H}{\to} X$ is smooth! But for your definition you would need that the composite of every non-smooth function with a morphism of test domains is still non-smooth.
See what I mean?
But what you could do is to take some set of maps that you want not to be smooth and then consider the largest diffeology (or whatever) such that those maps are not smooth. It’s possible that this will be the discrete diffeology (only the constant maps smooth), and there are one or two conditions that have to be satisfied to be sure of getting anything at all (you could never have a constant map not-smooth, for example).
On topology, you can certainly define a topology on a space by declaring a family of maps to be continuous. For example, that’s how you put a topology on a diffeological space (or Chen space or Frolicher space). Indeed, if you look at the topological sub-page of Froelicher spaces then you’ll see that I put two topologies on a Froelicher space: the curvaceous topology (strongest topology such that all smooth curves are continuous) and the functional topology (weakest topology such that all smooth functions are continuous).
This kind of construction is also used a lot in functional analysis: the weak and weak* topologies are defined by declaring certain families of functionals continuous. It’s also used in algebraic topology: the compactly-generated topology on a space is defined in this way.
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