# Meaning of “topologically equivalent”

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!

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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. –  Rudy the Reindeer 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. –  Rudy the Reindeer Jun 20 '11 at 11:12
It would be a very good exercise to make your intuition rigorous in this case. –  Dylan Moreland Jun 20 '11 at 12:23
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

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 $S^1 \to S^1$).
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