Is cone not a topological manifold?

Question

Is the cone = X a Hausdorff, second-countable topological space that is not a topological manifold? Since the open subsets $U_{\alpha}$ do not cover the vertex of the cone, so $U_{\alpha}$ is not a cover of X. Is this right? Thanks in advance :)

Answer 1

Hint: Consider your cone $X$ in $\mathbb{R}^3$ with a vertical axix. Then the projection $p : \mathbb{R}^3 \to \mathbb{R}^2 \times \{0\}$ gives you a homeomorphism $X \simeq \mathbb{R}^2$. Is $\mathbb{R}^2$ a topological manifold?