Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been wondering about "topologically equivalent" for some time now. For example:

$S^1$ is "topologically equivalent" to $\mathbb{R}P^1$.

I see that they are homotopy equivalent. But are they also homeomorphic? Probably yes.

Is there a failsafe way to determine whether in a given case "topologically equivalent" means "homeomorphic" or "homotopy equivalent"? Thanks for your help!

share|improve this question
1  
How can you see they're homotopy equivalent without seeing they're homeomorphic? –  Chris Eagle Jun 20 '11 at 10:29
    
I take $R^2$ and do the identification in my head. Then I end up with something that looks exactly like $S^1$, so both are a thing with one hole. It's a bit hand-wavy but I think if two spaces have the exact same shape then they are homotopy equivalent. –  Matt N. Jun 20 '11 at 10:51
    
I have seen "$=$" being used to mean either. Unfortunately, I can't remember where but I'm fairly sure it is not always used to mean homeomorphic. –  Matt N. Jun 20 '11 at 11:12
1  
It would be a very good exercise to make your intuition rigorous in this case. –  Dylan Moreland Jun 20 '11 at 12:23
1  
Or, using identifications, consider the upper-half of $S^1$, after the identifications have been made, you still need to identify the tips. You can do this with the map $e^it$ defined on the unit interval, which sends the points in the interior to interior point, and collapses 0 with 1. You then get the collection of points (cos2Pit,sin2Pit), t=0 to 1. –  gary Jun 20 '11 at 15:42

1 Answer 1

up vote 5 down vote accepted

It's ambiguous, and that's why you shouldn't use it. I've only seen the term used in popular texts to explain topology without explaining topology, and in my experience it could mean at least three things the way it's used:

  • homeomorphic
  • homotopy equivalent
  • isotopic.

In this case, $S^1$ and $\mathbb{R}P^1$ are homeomorphic. The explicit homeomorphism is not difficult to construct (it comes from the $2$-to-$1$ cover $S^1 \to S^1$).

share|improve this answer
    
Nice answer, thank you! –  Matt N. Jun 20 '11 at 12:06
    
I fixed a typo for $\mathbb{R}P^1$. (You wrote $\mathbb{R}P^2$.) I hope you don't mind. –  Zhen Lin Jun 20 '11 at 12:33
    
Hatcher does it : ( –  Matt N. Jun 24 '11 at 12:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.