# Exercises:Saul - Algebraic Topology - 1/Exercise 1.2/Lemmas

## Contents

## Exercises

### Exercise 1.2

#### Lemmas

**Caveat:**The following listed here are for reference to someone looking at the exercises only. They are done from memory and have no reference (at the time of writing) - use at your own peril.

##### Homeomorphisms and point removal

Suppose [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] are topological spaces and [ilmath]f:X\rightarrow Y[/ilmath] is a homeomorphism between them (so [ilmath]X\cong Y[/ilmath]), then for any [ilmath]x\in X[/ilmath] we have:

- [ilmath]f\vert_{X-\{x\} }:X-\{x\}\rightarrow Y-\{f(x)\} [/ilmath] is a homeomorphism
^{[Note 1]}we of course consider these spaces with the subspace topology

##### Point removal

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]f:X\rightarrow Y[/ilmath] be any map (possibly continuous) between them. Then

- [ilmath]f:X\rightarrow Y[/ilmath] being a homeomorphism
*implies*[ilmath]\forall p\in X[\mathcal{O}(p)\eq\mathcal{O}(f(p))][/ilmath] - where [ilmath]\mathcal{O}(p)[/ilmath] is the number of path-connected components of the space [ilmath]X-\{p\} [/ilmath]- We may write [ilmath]\mathcal{O}(p,X)[/ilmath] to mean [ilmath]p[/ilmath] removed from the space [ilmath]X[/ilmath], this makes things clearer when dealing with subsets.

Specifically, by contrapositive, if [ilmath]\exists p\in X[\mathcal{O}(p)\neq\mathcal{O}(f(p))][/ilmath] then [ilmath]f:X\rightarrow Y[/ilmath] is not a homeomorphism.

##### Two-step point removal

This is a corollary of the two claims above. Two-step point removal means that:

- Suppose [ilmath]X\cong_f Y[/ilmath], then [ilmath]X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\} [/ilmath] (by "
*homeomorphisms and point removal*") - Suppose [ilmath]X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\} [/ilmath] are indeed homeomorphic, then [ilmath]\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})][/ilmath]

We conclude:

- [ilmath]X\cong_f Y[/ilmath] implies [ilmath]\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})][/ilmath]

Which will be helpful for reaching contradictions.

## Notes

- ↑ This is a slight abuse of notation for a restriction, for a restriction we would have [ilmath]f\vert_{X-\{x\} }:X-\{x\}\rightarrow Y[/ilmath] - notice the codomain has changed to [ilmath]Y-\{f(x)\} [/ilmath]

## References