Set $n\geq 1$. Suppose $X\subset S^n$ is a neighborhood retract where $S^n$ is $n-$sphere. $X\subset Y$ is neighborhood retract if there is an open subset/neighborhood of $Y$ containing $X$ with $X$ as its retraction.
Denote $U$ as the neighborhood of $X$ as above for retraction. Then $H(U)=H(X)\oplus H(U,X)$. The claim is that $H_n(U)=0$. It suffices to look at $H_n(U,p)=\tilde{H}_n(U)=H_n(U)$ where $\tilde{H}$ is reduced homology and last step is by couple $(U,p)$ long exact sequence. WLOG $U$ is connected. Since $U$ is open, there is a even smaller open $B_p$ ball around $p$. Now $H_n(U,p)=H_n(U,B_p)$ where first step is by deformation retraction. How do I see $H_n(U,B_p)=0$?
$\textbf{Q:}$ I do not see how to proceed further as I do not have anything to cut off. I am being sloppy here. A lot of $=$ should be $\cong$.
$\textbf{Q':}$ The book says "because $X\neq S^n$, inclusion map $X\to S^n$ is nullhomotopic". How is so obvious? Am I missing something here?
Ref. Dold, Algebraic Topology Chpt IV, Sec 6, Cor 6.5 pg 73.
Best Answer
It is wrong if $X = S^n$ (it is well-known that the identity on $S^n$ is not nullhomotopic).
If $X \subsetneqq S^n$, choose any point $x \in S^n \setminus X$. Then $R = S^n \setminus \{ x \}$ is homeomorphic to $\mathbb{R}^n$ which is contractible, therefore the inclusion $X \to R$ is nullhomotopic. Hence also the inclusion $X \to S^n$ is nullhomotopic.
Note that this is true for any $X \subsetneqq S^n$.
Added:
For any open $U \subset S^n$ we have $H_i(S^n,U) = 0$ for $i > n$. This is is a special case of Proposition 6.4 in Dold. The proof is not at all trivial. It begins on p. 73 (Lemma 6.7) and ends on p.77.
Accepting $H_i(S^n,U) = 0$ for $i > n$, we see that the long exact sequence of $(S^n,U)$ contains the part $H_{n+1}(S^n,U) \to H_n(U) \to H_n(S^n)$. Since the first group is $0$, we see that $i_* : H_n(U) \to H_n(S^n)$ is a monomorphism. In particular $H_n(U) \approx \text{im}(i_*)$. But if $U \subsetneqq S^n$, then the inclusion $i : U \to S^n$ is nullhomotopic. Hence $i_*$ is the zero homomorphism and therefore $\text{im}(i_*) = 0$. We conclude $H_n(U) = 0$.
Note that the idea sketched in your question does not work. It is really hard work to prove $H_n(U) = 0$.