Open sets in identification topology.

Question

I'm reading about identification space and I think I misunderstood some point in the excerpt below.

The part in red says that the image of the open half-disc of $R$ under $\pi$ is not open in $M$, but I can't see why (or at least why the following reasoning doesn't hold). Since $R \subset \mathbb{E}^2$, I assume the half-disc is open in the subspace topology, and it is the intersection of $R$ with an open set of $\mathbb{E}^2$? Then isn't it the case that the image of this open half-disc has as pre-image an open set of $R$, and then by the definition of the identification topology, that image should be open?

On a side note, if we see $M$ as a subspace of $\mathbb{E}^3$, then can the image of the open half-disc be seen as the intersection of $M$ with an open ball in $\mathbb{E}^3$, and hence open? Is this correct and relevant to our discussion, or am I talking about a whole different topology?

Apologize if this seems trivial. I just want to root out all the misunderstanding that I have.

Answer 1

Your error is in the following sentence:

Then isn't it the case that the image of this open half-disc has as pre-image an open set of R, and then by the definition of the identification topology, that image should be open?

Because the pre-image of the image of the half-disc is not the half-disc, since $\pi$ is not injective. It's the half disc plus a bit of the boundary line on the other side of $R$. And that set is not open in $R$.

Answer 2

While the half-disc $D$ is open in $R$, its image $\pi(D)$ is not open in $M$. To see why, consider any small neighborhood $V$ of $\mathbf{p}$ . $\pi^{-1}(V)$ is the disjoint union of a neighborhood of $(0,y)$ and a neighborhood of $(3,1-y)$, hence $V$ cannot be contained in $\pi(D)$.