# B^n/S^{n-1} is homeomorphic to S^n

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Please check whether the following statement is true:

Lemma:Let $X$ be any Hausdorff space. Let $U$ be open in $X$. Let $y \in U$. Then $U \setminus \{ y \}$ is open in $X$.

Proof: Write $V=U \setminus \{ a \}$. Let $x \in V$. Then $\exists$ open sets $A$ and $B$ of $X$ such that $x \in A , y \in B$ and $A \cap B = \emptyset$. Then $A \cap U$ is an open in $X$ which is a subset of $V$ and contains $x$. This can be done for all $x \in V$ thus $V$ is open.

Assuming the above is fine, I need to prove this:

Let $p$ be a point in $S^n$.
Let $h_1 : B^n - S^{n-1} \rightarrow \mathbb{R}^n$ and $h_2 : \mathbb{R}^n \rightarrow S^n - \{ p \}$ be homeomorphisms.
Define $f : B^n \rightarrow S^n$ as:
$f(x)=h_2 h_1 (x) \, \, \text{ for } x \in B^n - S^{n-1}, f(x)=p \, \, \text{ for } x \in S^{n-1}$.
Show that $f$ is continuous.

Attempt:
Let $U$ be open in $S^n$. If $p \notin U$ then its easy to show that $f^{-1}(U)$ is open in $B^n$.
Here's the problem case. Assume $p \in U$. Clearly $S^n$ is Hausdorff, so by the statement proved above $U- \{ p \}$ is open in $S^n$.
So $f^{-1}(U)=f^{-1}(U-\{ p \}) \cup f^{-1}(p)= \text{ (an open set in B^n)} \cup S^{n-1}$.
If $S^{n-1}$ were open in $B^n$ I'd be done, but it ain't. What do I do?

NOTE: $B^n$ is the unit ball in $\mathbb{R}^n$ and $S^{n-1}$ is its boundary.

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#### Opalg

##### MHB Oldtimer
Staff member
Please check whether the following statement is true:

Lemma:Let $X$ be any Hausdorff space. Let $U$ be open in $X$. Let $y \in U$. Then $U \setminus \{ y \}$ is open in $X$.

Proof: Write $V=U \setminus \{ a \}$. Let $x \in V$. Then $\exists$ open sets $A$ and $B$ of $X$ such that $x \in A , y \in B$ and $A \cap B = \emptyset$. Then $A \cap U$ is an open in $X$ which is a subset of $V$ and contains $x$. This can be done for all $x \in V$ thus $V$ is open.
Yes, that looks fine.

Assuming the above is fine, I need to prove this:

Let $p$ be a point in $S^n$.
Let $h_1 : B^n - S^{n-1} \rightarrow \mathbb{R}^n$ and $h_2 : \mathbb{R}^n \rightarrow S^n - \{ p \}$ be homeomorphisms.
Define $f : B^n \rightarrow S^n$ as:
$f(x)=h_2 h_1 (x) \, \, \text{ for } x \in B^n - S^{n-1}, f(x)=p \, \, \text{ for } x \in S^{n-1}$.
Show that $f$ is continuous.

Attempt:
Let $U$ be open in $S^n$. If $p \notin U$ then its easy to show that $f^{-1}(U)$ is open in $B^n$.
Here's the problem case. Assume $p \in U$. Clearly $S^n$ is Hausdorff, so by the statement proved above $U- \{ p \}$ is open in $S^n$.
So $f^{-1}(U)=f^{-1}(U-\{ p \}) \cup f^{-1}(p)= (\text{an open set in } B^n) \cup S^{n-1}$.
If $S^{n-1}$ were open in $B^n$ I'd be done, but it ain't. What do I do?

NOTE: $B^n$ is the unit ball in $\mathbb{R}^n$ and $S^{n-1}$ is its boundary.
Can you get round this by looking at the complements of the sets? If $U$ is an open set containing $p$, then the complement $K$ of $U$ in $S^n$ is a compact set not containing $p$. Since $h_2$ and $h_1$ are homeomorphisms it follows that $h_2^{-1}(K)$ is a compact subset of $\mathbb{R}^n$, and $h_1^{-1}\bigl(h_2^{-1}(K)\bigr)$ is a compact subset of $B^n - S^{n-1}$. Deduce that its complement in $B^n$, which is equal to $f^{-1}(U)$, is an open set containing $S^{n-1}.$

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Yes, that looks fine.

Can you get round this by looking at the complements of the sets? If $U$ is an open set containing $p$, then the complement $K$ of $U$ in $S^n$ is a compact set not containing $p$. Since $h_2$ and $h_1$ are homeomorphisms it follows that $h_2^{-1}(K)$ is a compact subset of $\mathbb{R}^n$, and $h_1^{-1}\bigl(h_2^{-1}(K)\bigr)$ is a compact subset of $B^n - S^{n-1}$. Deduce that its complement in $B^n$, which is equal to $f^{-1}(U)$, is an open set containing $S^{n-1}.$
Thank You Opalg. Your posts help me so much that I can't thank you enough.