$X \times Z \cong Y \times Z \implies X \cong Y$?

Question

Let $X,Y,Z$ be topological spaces. Is the following statement true? $X \times Z \cong Y \times Z \implies X \cong Y$? how would you prove it?

and I know that if $A \cong B$, and $a \in A$ that there is a $b \in B$, such that $A\setminus{\{a\}} \cong B\setminus{\{b\}}.$ How would you prove the same for removing lines from product topology, instead of point of normal topological spaces?

Answer 1

In addition to the counterexamples given in Henno Brandsma's answer, there is a very pathological and very interesting counterexample: In 'A counterexample related to topological sums – Yamamoto, Shuji and Yamashita, Atsushi' a space $X$ and a space $Y$ are constructed such that $X$ and $Y$ are not homeomorphic but $X\times 2$ and $Y\times 2$ are. (By $2$ it is meant the discrete space with $2$ elements.) Notice that $X\times 2$ is just two copies of $X$, side by side, topologically unrelated to each other. Also, amazingly, $X$ and $Y$ are compact metric spaces—in fact, subspaces of $\mathbb{R}^2.$

This relates to your question in that $2\times X \cong 2\times Y$ but $X\not\cong Y$.

Also, much simpler albeit much less interesting counterexamples can be obtained with discrete spaces, e.g., $\mathbb{Z}\times 2\cong \mathbb{Z}\times 3$ but $2\not\cong 3$.

Answer 2

Not true. Some examples, let C be the Cantor set :

$C \times C \simeq C \times \{0\}$ but $C \not\simeq \{0\}$.

$\mathbb{Q} \times \mathbb{Q} \simeq \mathbb{Q} \times \mathbb{Z}$ but $\mathbb{Q} \not\simeq \mathbb{Z}$

$[0,1] \times [0,1]^{\mathbb{N}} \simeq [0,1]^2 \times [0,1]^{\mathbb{N}}$ but $[0,1] \not\simeq [0,1]^2$.

$[0,1) \times [0,1) \simeq [0,1) \times [0,1]$ etc.

Answer 3

Let $ X=\mathbb{R^+} \cup \{0\} $ then $ \mathbb{R} \times X \cong X\times X $ with homeomorphism $(x,y) \to (x^2 -y^2,2xy)$ but $ \mathbb{R^+} \cup \{0\} \ncong \mathbb{R} $