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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am confused by the orientation of a topological manifold. My understanding is: An orientation of a topological manifold is a choice of generator of the $H^n(M,\mathbb Z)$. So given a manifold, we could have 2 orientation defined on the manifold. For $\mathbb{CP}^2$ on orientation is determined by the complex structure, the other orientation is denoted by $\overline{\mathbb{CP}}^2$. And it is well known that there is no orientation reversing map from $\mathbb{CP}^2$ to itself. My question is: are they homeomorphic? I guess they are not homeomorphic. But I am confuesed, doesn't the orientations defined on the same manifold, how come after reversing the orientation they become not homeomorphic?

share|cite|improve this question
I don't believe a homeomorphism needs to preserve orientation. It would, however, induce an isomorphism on cohomology. – alex.jordan Apr 13 '12 at 2:09
It's true homeomorphisms are not required to preserve orientation. – Grumpy Parsnip Apr 13 '12 at 2:13
Thanks, But whether $\mathbb{CP}^2$ and $\overline{\mathbb{CP}}^2$ homeomorphic to each other? – user17150 Apr 13 '12 at 2:20
It seems the identity map should be a homeomorphism. – M Turgeon Apr 13 '12 at 2:26
But there is no orientation reversing map from $\mathbb{CP}^2$ to itself. which implies the above two are not homeomorphic! I am crazy.... There must be some stupid mistake I made in my argument, but i can't find it. – user17150 Apr 13 '12 at 2:31
up vote 3 down vote accepted

A topological manifold is a topological space with extra conditions. In particular, for $\mathbb{CP}^2$, whatever the orientation you choose, the underlying topological space is the same. Therefore, the identity map is a homeomorphism.

However, they are not equivalent as oriented topological manifold, since there exists no orientation-reversing map from $\mathbb{CP}^2$ to itself. Hence, I think the confusion arises from a confusion with terminology.

share|cite|improve this answer
So the idendity map between $\mathbb{CP}^2$ and $\overline{\mathbb{CP}}^2$ can not be considered as a homeomorphism between these two 'oriented' manifolds. Correct? – user17150 Apr 13 '12 at 2:46
Well, usually, I don't see the word homeomorphism being used as an orientation-preserving isomorphism. Homeomorphism means that the topological structure is preserved, and certainly the identity map is a homeomorphism. The orientation is an extra structure, and the identity map does not preserve it. To summarize, these two manifolds are homeomorphic, but not in an orientation-preserving way. – M Turgeon Apr 13 '12 at 2:53

As M Turgeon points out, the issue is one of labeling. I'm writing this as an answer just to be a little more precise with notation to try to be absolutely certain the confusion is gone.

Let's call the underlying topological manifold $X$. (Just to be concrete, let's define $X=e^0\cup e^2\cup e^4$, the usual cellular decomposition.) Now, let $a$ and $b$ denote the two possible generators of $H_n(X;\mathbb{Z})$. We define the oriented manifolds $\mathbb{CP}^2 = (X,a)$ and $\overline{\mathbb{CP}^2} = (X,b)$. Note that these are not just topological spaces, they're topological spaces each with a different additional choice made. Thus, they're represented as pairs.

The identity map $\iota:X\to X$ induces a homeomorphism $\iota:\mathbb{CP}^2\to\overline{\mathbb{CP}^2}$ when we add the orientation data. In fact, this homeomorphism has $\iota_*a = a = -b$, so it does not reverse orientation. (Think about that for a moment. To reverse orientation, a homeomorphism $h$ of $X$ must carry $a$ to $b$. If the map instead carries $a$ to $-b = a$, it preserves orientation.)

To be a little more precise, it may be useful to extend the notion of homeomorphism to isomorphism of oriented manifolds. An isomorphism of oriented manifolds $i:(Y,\alpha)\to (Z,\beta)$ is a homeomorphism of the underlying topological spaces which carries the marked generator $\alpha$ of $H_n(Y)$ to the generator $\beta$ of $H_n(Z)$.

Note the linguistic trickery here: an orientation-reversing homeomorphism of $X$ is an isomorphism of the oriented spaces $\mathbb{CP}^2$ and $\overline{\mathbb{CP}^2}$. Now we see that the statement "$\mathbb{CP}^2$ has no orientation-reversing homeomorphism" unpacks to:

There is no isomorphism of oriented manifolds between $\mathbb{CP}^2$ and $\overline{\mathbb{CP^2}}$, that is, there exists no homeomorphism $h:X\to X$ such that $h_*a = b$.

Exercise: Distinguish between an orientable topological space and an oriented topological space. (If general topological spaces are too subtle, work with manifolds.)

Exercise: I assumed you know what an "oriented manifold" is. Make a precise definition.

share|cite|improve this answer
Thanks, that's really help. – user17150 Apr 13 '12 at 3:19
You're welcome! – Neal Apr 13 '12 at 14:16

Your Answer


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.