I'm following an online course on Differential Geometry by F. Schuller and in this lesson https://www.youtube.com/watch?v=93f-ayezCqE&t, around 23:00, he draws a bifurcation and says that isn't a topological manifold because any open set around the bifurcation can't be mapped continuously and with continuous inverse to $\mathbb{R}$, $\mathbb{R}^2$, $\mathbb{R}^3$, etc.
I don't understand why. If we take $\mathbb{R}^2$, I can take every point of the bifurcation and map it to a curve in $\mathbb{R}^2$ (so, a section of $\mathbb{R}^2$) that looks like a Y. Similarly, I can take any point of that curve and map it back to the manifold. What is the problem here? I get why it isn't a differentiable manifold, but why not a topological manifold?
EDIT: Solved. The problem was that I missed that, by definition, in each point the neighborhood must be mapped not only to a subset of $\mathbb{R}^2$, but this subset must be open as well, so it can't be a curve in $\mathbb{R}^2$.
Intuitively, any neighborhood of the triple point looks like a $Y$. If this is supposed to have a dimension, it should be $1$-dimensional, since everything away from the triple point looks like an arc. But the triple point itself is not homeomorphic to any connected open set in $\mathbb{R}$. Indeed, if we remove the triple point from our neighborhood, we are left with three connected components (the legs of the $Y$), but for any open interval $(a,b) \subseteq \mathbb{R}$ if we remove a point we're left with two components! So there can be no homeomorphism from any neighborhood of the triple point to any subset of $\mathbb{R}$, and so the figure is not a manifold.
There are ways to make precise the intuitive idea that if this were a manifold, it would have to be $1$-dimensional (for instance, using invariance of domain) which I'll leave unsaid here for simplicity.
I hope this helps ^_^