I want to show that a the standard 2-Simplex in $\mathbb{R^2}$, i.e. the set of all convex combinations of the three vectors $0, e_1, e_2$ can be viewed as a manifold with corners.
However, I am already stuck showing the locally euclidean property around the point $e_1$, i.e. given the point $(1,0)$. I wanted to find a open set around it (in the subspace topology) which is homemorphic to a relatively open set of $\mathbb{R_{+}^2}$. A natural candidate map would be: $$\phi: (x_1,x_2) \mapsto (1-x_1,1-x_1-x_2)$$ In this case the point $e_1$ would be a corner point as it is mapped to $(0,0)$. However, I think that the given map is not a homeomorphism (reasoning given below). I would be glad if someone could either point out that my reasoning is wrong and that the map is a homeomorphism or - if the reasoning is correct - provide me with a homeomorphism which maps $e_1$ to $(0,0)$.
My reasoning: The inverse map $\phi^{-1}$ is given by: $$\phi^{-1}: (y_1,y_2) \mapsto (1-y_1, y_1-y_2)$$ Now for the image to be contained in the simplex it is necessary that $y_1-y_2\geq 0$ holds. But then around the point $y_1=0, y_2=0$ no choice of any neighborhood (relative to $\mathbb{R_{+}^2}$) would fulfill that each point in it still satisfies this inequality.