Continuity and non-surjective functions

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Consider a path in a topological space, [ilmath](X,\mathcal{J})[/ilmath], then a path is simply:

  • [ilmath]p:([0,1]\subset\mathbb{R},\vert\cdot\vert)\rightarrow(X,\mathcal{J})[/ilmath] where [ilmath]\vert\cdot\vert[/ilmath] denotes the absolute value

Then for this to be continuous we require that:

  • [ilmath]\forall U\in\mathcal{J}[p^{-1}(U)\in\mathcal{O}([0,1],\vert\cdot\vert)][/ilmath] where [ilmath]\mathcal{O} [/ilmath] denotes the open sets of a space


I was once puzzled by this definition

TODO: Add picture

For consider a path on a surface, and now take any open set that intersects with that path. A good chunk of that isn't used, what if it overlaps with 2 or more line segments?

  • None of this is a problem, even if the open set doesn't intersect with the line at all, well [ilmath]p^{-1}(U)=\emptyset[/ilmath], but the empty set is open in any topology, so it isn't a problem.
  • Topologies are also closed under unions (any union) - so even if there are multiple parts of the loop, it doesn't matter.

It didn't take me long to realise this but it's important to understand what it means when we throw continuity around like this.

Possible theorem

TODO: Attempt proof, I'm on the train right now!

  • A map is continuous [ilmath]\iff[/ilmath] it is continuous with the subspace topology on its image

Notes/Work in progress

This is a WIP section

Let [ilmath]f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})[/ilmath] be a non-surjective map

I claim that this is continuous [ilmath]\iff[/ilmath] [ilmath]f':(X,\mathcal{J})\rightarrow(I,\mathcal{S})[/ilmath] is continuous where:
  • [ilmath]I:=\text{Im}(f)[/ilmath] and
  • [ilmath]\mathcal{S}[/ilmath] is the subspace topology on [ilmath]I[/ilmath] induced by [ilmath]\mathcal{K} [/ilmath]

Proof: [ilmath]f[/ilmath] is continuous [ilmath]\implies[/ilmath] [ilmath]f'[/ilmath] is continuous

Let [ilmath]V\in\mathcal{S} [/ilmath] be given, we wish to show that [ilmath]f'^{-1}(V) [/ilmath] is open in [ilmath]X[/ilmath]
By hypothesis [ilmath]\exists U\in\mathcal{K} [/ilmath] such that [ilmath]f^{-1}(U) [/ilmath] is open, and additionally, [ilmath]V=U\cap I[/ilmath]
We need to show that [ilmath]f'^{-1}(V)=f^{-1}(U)[/ilmath]
Proof that [ilmath]f'^{-1}(V)\subseteq f^{-1}(U)[/ilmath]
Let [ilmath]x\in f'^{-1}(V)[/ilmath], then [ilmath]\exists y\in V[/ilmath] such that [ilmath]f'(x)=y[/ilmath]
As the intersection of sets is a subset of each set and [ilmath]V=U\cap I[/ilmath] we have:
  • [ilmath]V\subseteq U[/ilmath] - which is useful immediately, and
  • [ilmath]V\subseteq I[/ilmath] - which isn't useful yet..
Using [ilmath]V\subseteq U[/ilmath] by the implies-subset relation we see that
  • [ilmath]y\in V[/ilmath] and [ilmath]y\in I[/ilmath]
we have [ilmath]x\in f'^{-1}(V)\implies\exists y\in V[f'(x)=y\wedge y\in U\wedge y\in I][/ilmath]
as [ilmath]y\in I[/ilmath] we know [ilmath]\forall A[y\in A\implies f^{-1}(A)\ne\emptyset][/ilmath] - for any set containing [ilmath]I[/ilmath] the inverse isn't empty
  • As by definition, [ilmath]I[/ilmath] is the image!
Thus [ilmath]\implies\exists x'[f^{-1}(y)=x'][/ilmath]
As [ilmath]y\in U[/ilmath] we know that [ilmath]x'\in f^{-1}(U)[/ilmath]

TODO: Show equality?