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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!

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


However, notice that a bijective continuous map is not necessarily a homeomorphism.

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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.

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Intuitively: an homomorphism is a correspondence that guaranties the equivalence from two topological spaces, that are two sets, each with a family of subset, callet the topology, and whosw elements are the open sets. This means that not only the two sets have to be equivalent (that is guarantied by the fact that the correspondence is a bijection) bat that also the two topologies are ''equivalent'' that is that the open sets of one topology are the image of the open sets of the other, and thsi is guarantied by the request that the correspondence is continuous in both verses.

The answer of Daniel illustrate how we can define a topology such that a bijection be continuousso that it becomes an homomorphism.

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