Let $(X, \tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A bijection $\gamma:X\to Y$ is called a homeomorphism if:

  1. $\gamma$ is continuous, and
  2. $\gamma$ has some continuous inverse $\gamma^{-1}:Y\to X$.

Also recall that $\gamma$ is (topologically) continuous if the preimage of each open set $U\in\tau_Y$ is also open. Given that $\gamma^{-1}$ is also required to be continuous, we have that $\gamma$ induces a bijection on the topologies $\tau_X$ and $\tau_Y$.

My question is: why do we need the initial bijection condition? Given a map $\phi:X\to Y$ which induces some bijection $\tau_X\to\tau_Y$, why isn't $\phi$ a homomorphism?

  • $\begingroup$ Consider $\Bbb R$ with the topology where $\Bbb R$ and $\emptyset$ are the only open sets, and map $\phi : \Bbb R\to\{\ast\}$ (map everything to $\ast$), where $\{\ast\}$ is given the topology where $\{\ast\}$ and $\emptyset$ are the only open sets. Then the map $\phi$ maps $\Bbb R$ to $\{\ast\}$ and $\emptyset$ to $\emptyset$, and $\phi^{-1}$ sends $\{\ast\}$ to $\Bbb R$ and $\emptyset$ to $\emptyset$, but $\phi$ has no inverse as a map of sets. $\endgroup$
    – Stahl
    Oct 25, 2016 at 17:03

1 Answer 1


Let $\langle X,\tau\rangle$ be any space. Let $D=\{0,1\}$ with the indiscrete topology $\{\varnothing,D\}$, and let $Y=X\times D$ with the product topology $\tau_Y$. The projection map $\pi_X:Y\to X:\langle x,d\rangle\mapsto x$ induces a bijection from $\tau_X$ to $\tau_Y$, but it’s not a homeomorphism. Specifically, the induced bijection sends $U\in\tau_X$ to $U\times D\in\tau_Y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.