Local homology group: a homeomorphism takes the boundary to the boundary 
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.
 A: Yes, you have the right idea, but you seem confused with some definitions. The proof goes as follows: Denote by $X$ the euclidean half-space of dimension $n$ and by $Y$ its boundary. You might find useful this lemma, which is a direct consequence of the excision theorem
Lemma. Let $A\subset X$. If $A$ contains a neighborhood of the point $x\in X$ then
$$
H_p(X,X - x) \cong H_p(A,A - x)
$$
So now let's prove something stronger than your part a) which will help to get part b) as you wanted.
Proposition. Let $x\in X$ then
$$
H_i(X,X- x) = 
\begin{cases}
\mathbb{Z} & \text{if } x\notin Y \text{ and } i=n\\
0 & \text{otherwise}
\end{cases}
$$
Proof. Suppose that $x\notin Y$, then we can find a neighborhood $U$ of $x$ homeomorphic to $\mathbb{R}^n$ and by the preceding lemma we have that 
$$H_i(X,X - x) \cong H_i(\mathbb{R}^n,\mathbb{R}^n - x) \cong H_i(B^n,B^n - x)$$
Using that $S^{n-1}$ is a deformation retract of $B^n$ we have that $H_i(B^n,B^n - x) \cong H_i(B^n,S^{n-1})$ by the exact sequence of the pair we have that this group is $\mathbb{Z}$ if $i=n$ and $0$ otherwise. Now  suppose that $x\in Y$, composing with a homeomorphism we can assume that $x= 0$. denote by $B_+^n$ the half ball in $X$, i.e. $B^n\cap X$ then we have by the previous lemma that $$H_i(X,X-x) \cong H_i(B_+^n,B_+^n-x)$$
Now there is a deformation retraction of $B^n - x$ onto $S^{n-1}$ that restricts to $B_+^n-x$ onto $S_+^{n-1}$. Using that $S_+^{n-1}\cong B^{n-1}$ (draw a picture) we have by the long exact sequence of the pair that their homology vanishes.
Now for part (b) suppose that you have an homeomorphism $h: X\to X$ that not preserve the image of $Y$, then choose a point $y\in Y$ such that $f(y)\notin Y$, by the previous proposition their local homology groups are not isomorphic, contradiction.
