Continuity and non-surjective functions

Basis

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

Misconception

I was once puzzled by this definition

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TODO: Add picture

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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

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

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