Topological equivalence

In my understanding, a bijective map establishes a one-to-one correspondence between elements, and has an inverse. This represents that the map garantees topologically equivalent transformation. (*homeomorphism == bijective, continuous inverse)

If a topological structure is transformed(mapped) with a bijective function to the other topological space, then the transformed structure is topologically equivalent with the original one like donut == mug cup.

I just want to (intuitively) understand how the bijective map garantees the topological equivalence in topological structure mapping. How do I know that the transformed structure is topologically equivalent with the original one?

How does the one-to-one mapping garantee the topological equivalence, conceptually?

Thanks.

Thanks!

• Because it preserves the topology i.e. the two spaces have the same open sets as indicated in Daniel's answer. – Vim Mar 29 '18 at 8:47

Suppose $X$ is a topological space, and $Y$ is a set, and let $f:X\to Y$ be a bijective map of sets. Then you can endow $Y$ with the finest topology such that $f$ is continuous, which is given by $$O\subseteq Y\text{ open in }Y\iff f^{-1}(O)\text{ open in }X\ .$$ Then, since $f$ is bijective, if $U\subseteq X$ is open, we have $f^{-1}f(U) = U$, so that $f(U)$ is open, which implies that $f$ is a homeomorphism.
It doesn't. Take the identity function from $\mathbb R$ endowed with the discrete topology into $\mathbb R$ endowed with the usual topology. Then this function is a continuous bijection, but its inverse is not continuous.