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Background:

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?

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  • $\begingroup$ 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 '16 at 17:03
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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 $U\times D\in\tau_Y$.

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