Orientation on $\mathbb{CP}^2$

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?

Solution 1:

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.

Solution 2:

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.