So I need to find finite topological spaces $X,Y$ and a map $f:X \rightarrow Y$ which is a continuous bijection but not a homeomorphism.

My attempt: Let both $X$ and $Y$ be the set $\{a,b,c\}$, and let $f$ be the identity function. Let $X$ have the discrete topology and let $Y$ have the indiscrete topology. Now, $f$ is clearly a bijection, and the pre-images are $f^{-1}(\emptyset) = \emptyset$ and $f^{-1}(Y) = X$. So $f$ is continuous.

Now, consider $f^{-1}$. The pre-image $(f^{-1})^{-1}(a) = f(a) = a$, but while $a$ is open in $X$, it's not open in $Y$. Therefore $f^{-1}$ is not continuous, and so we are done.

Is this a correct example?

  • 2
    $\begingroup$ Yes, it’s correct. You could also use the two-point discrete and indiscrete space, or the two-point discrete space and the Sierpiński space, among many other examples. $\endgroup$ – Brian M. Scott Apr 20 '16 at 11:33
  • $\begingroup$ Yes, in fact this example works for sets of any cardinality greater than $2$. $\endgroup$ – sqtrat Apr 20 '16 at 11:33
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    $\begingroup$ More generally the identity function $(X,\tau_1)\to (X,\tau_2)$ is continuous if and only if $\tau_2\subset \tau_1$. In particular it cannot be an homeomorphism if the two topologies $\tau_1,\tau_2$ are different. $\endgroup$ – Arnaud D. Apr 20 '16 at 11:41

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