Let $X \subset \mathbb{R}^{n}$ be the subspace $\{ x_{1},...,x_{n} \mid x_{n}\geq 0 \}$, and let $Y$ is the subspace with $x_{n}=0$.
Let $x\in Y$, calculate the local homology of $X$ at $x$.
Prove that any homeomorphism $h:X\to X$ must take $Y$ to $Y$.
For part 1, I know that if I can find a closed contractible neighborhood of $x$, and the boundary of $N$ is $B$, $B$ is the deformation retraction of $N-{x}$ then $H_{n}=H_{n-1}(B)$ the second I mean the reduced homology group.
But since $x$ is on the boundary of $X$, so I wonder if I can still use this, and for the easiest case, which we consider the upper half plane and a point on the $X$-axle , then the complement of a point is contractible, so I think that the homology group should be $\mathbb{Z}$ for $n=0$, and $0$ otherwise. But I am not sure if that is right and I still want the details that I can use to solve this part 1.
For part 2, $Y$ is the connected compact subset, so the homeomorphism preserves the image of $Y$ to be the connected and compact subset, if the image of $Y$ is not $Y$, then we can choose a point in $X$ that is not on the boundary of $X$ then the local homology of $X$ at $x$ and $f(x)$ are different, contradicts. I wonder if this is the right method.