Is it true that $X\simeq S^2\vee S^2$?

Question

Let $X$ be the quotient space of $S^2$ under the identifications $x\sim -x$ for every $x$ in the equator $S^1$. Is it true that $X\simeq S^2\vee S^2$, that is, $X$ is homeomorphic to $S^2\vee S^2$?

Answer 1

You can consider the cellular homology with $\Bbbk=\Bbb Z/2\Bbb Z$-coefficients. Both spaces are CW complexes, the quotient space $X=S^2/\sim$ has a CW complex structure with one $0$-cell, one $1$-cell and two $2$-cells attached to the one skeleton (a circle) by degree $2$ maps, while the wedge sum $Y=S^2\vee S^2$ has a CW structure with one $0$ cell and two $2$ cells. Their cellular homology (with $\Bbb Z/2\Bbb Z$-coefficients) is the homology of the complex $$0\to \Bbbk\oplus\Bbbk\xrightarrow{0}\Bbbk\xrightarrow{0}\Bbbk\to 0$$ for $X$, and $$0\to \Bbbk\oplus\Bbbk\to0\to\Bbbk\to 0$$ fro $Y$. The differentials are all $0$ (obvious for the second one, and follows from the degree $2$ remark above for the first one), so that the complexes are already the homology. Since they are different, the two spaces cannot be homeomorphic or even homotopy equivalent.

Answer 2

No, it is not true. On $S^2 \vee S^2$ there is a unique point which has no neighborhood homeomorphic to $\mathbb{R}^2$. On $X$, there are infinitely many such points.