Proof verification : $D^2 = \{(x,y)\in \Bbb R^2 : x^2 + y^2 \leq1\}$ is not homeomorphic to $\{-1,1\}\times D^2$ I am trying to prove that $D^2 = \{(x,y)\in \Bbb R^2 : x^2 + y^2 \leq1\}$ is not homeomorphic to $\{-1,1\}\times D^2$ .
My attempt:
$D^2$ is path connected and hence connected ( given any $x,y \in D^2$, define $\gamma : [0,1] \to D^2$ by, $\gamma(t)= (1-t)x+ty$, then $\gamma(0)=x,\gamma(1)=y$ and $|\gamma(t)| = |(1-t)x+ty)|\leq |1-t||x| +|t||y|\leq 1-t+t=1 \forall 0\leq t \leq 1$ )
While we can define $p:\{-1,1\}\times D^2 \to \{0,1\}$ by, $p(x) = 0  \text{ if } x \in \{-1\} \times D^2$ and $p(x) =1   \text{ if } x \in \{1\} \times D^2$ . And $p$ turns out to be  Continuous, thus $\{-1,1\}\times D^2$ is disconnected.
Is my argument correct?
 A: First and the most important question is: what are topologies on $D^2$ and $\{-1,1\}$? You've completely ignored that. And note that with discrete topologies $D^2$ is indeed homeomorphic to $\{-1,1\}\times D^2$.
So I will assume that you are refering to Euclidean topology in both cases. In particular $\{-1,1\}$ remains discrete.
In that case your argument is valid, although you are making things more complicated then necessary. In particular you would have to prove that $p$ is continuous.

Lemma. Let $X$ be a non-empty topological space and $A$ a disconnected space. Then $X\times A$ is disconnected.

Proof. Since $A$ is disconnected then $A=U\cup V$ for some open subsets $U,V\subseteq A$ which are nonempty and disjoint. Then
$$(X\times U)\cup (X\times V) = X\times A$$
and note that both $X\times U$ and $X\times V$ are open (by the definition of product topology). They are also nonempty and disjoint. This completes the proof. $\Box$
So in particular $\{-1,1\}\times D^2$ is disconnected because $\{-1,1\}$ (with discrete topology) is. On the other hand $D^2$ (with Euclidan topology) is connected as you've pointed out. Hence they cannot be homeomorphic.
A: The general approach is correct: $D^2$ is connected, but $\{1,-1\}\times D^2$ isn't. But in the second paragraph you should have explaind why $\{1,-1\}\times D^2$ is disconnected. You should have added that if it was connected, then, since $P$ is continuous, $p(\{1,-1\}\times D^2)$ would be connected, too.
And, although it is correct to use the space $\{0,1\}$ here, it would have been more natural to use $\{1,-1\}$ instead.
