Continuity of $(x,y)\mapsto(y,x)$

Question

Is this map continuous for real spaces $X,Y$? I can’t see why it wouldn’t be but at the same time I can’t seem to come up with concrete reasoning why it would be true. Thanks in advance!

Answer 1

Yes, it's true for $\mathbb{R}\times \mathbb{R}$ (and more generally for product topologies). You can prove it by reasoning through the definition of continuity in terms of open sets, and the inverse images under $f:\langle x,y\rangle \mapsto \langle y, x\rangle.$

• There's an open box $B = (a_1, a_2) \times (b_1, b_2)$ which is a product of open intervals.

• In other words, $B$ consists of all points $\{ \langle x,y \rangle : a_1 < x < a_2; b_1 < y < b_2\}$.

• Consider the map $f:\langle x,y\rangle \mapsto \langle y,x\rangle$. What do inverse images under $f$ look like? You can convince yourself that $f^{-1}(B)$ is the set of all points in $B$ with their coordinates reversed, namely: $$B^\prime = (b_1,b_2)\times (a_1, a_2)$$

• In $\mathbb{R}^2$ with its usual topology, the open boxes are all open. So $B^{\prime}$ is open. We have shown that the inverse image of every open box is open.

• But $\mathbb{R}^2$ additionally has a more special property, which is that (in its usual topology) every open set can be built up as a union of open boxes. But then for any collection of open boxes $\{B_\alpha\}$, we've got:

  $$f^{-1}\left(\bigcup_\alpha B_\alpha\right) = \bigcup_{\alpha} f^{-1}(B_\alpha) = \bigcup B_\alpha^\prime.$$

  (The first equality is just a true statement about how inverse images work.)

  So the inverse image of every open set is open, proving that $f$ is continuous.

Answer 2

If you use the universal property of product topology, it is immediate that $f(x,y) = (y,x)$ is continuous because$p_1\circ f $ and $p_2 \circ f$ are continuous, where $p_1,p_2$ are the natural projection to the first and second components respectively.