Two spaces are the same, topologically, when there is a bijection between their points that matches their open sets exactly. Such a map is a homeomorphism, and it is the right notion of sameness here: it is continuous in both directions, so nothing stated in open sets can tell the two spaces apart.

A cup and a doughnut are the same shape to a topologist because one can be moulded into the other without tearing or gluing. The moulding is the map; being able to mould back is what makes it an equivalence.

A homeomorphism matches the topologies: $U$ is open exactly when $f(U)$ is. Both maps are continuous, so open sets travel in both directions, and any statement made only about open sets holds in one space exactly when it holds in the other.

A homeomorphism is a bijection $f:X\to Y$ such that both $f$ and $f^{-1}$ are continuous. Two spaces are homeomorphic, written $X\cong Y$, when one exists. Being homeomorphic is an equivalence relation: the identity is a homeomorphism, an inverse of one is one, and a composition of two is one.

• A continuous bijection need not be a homeomorphism. The identity from $\mathbb{R}$ with the discrete topology to $\mathbb{R}$ with the standard one is a continuous bijection whose inverse is not continuous.
• A bijection is a homeomorphism exactly when it is continuous and open — that is, when it sends open sets to open sets. The theorem below is this statement.
• $(0,1)\cong\mathbb{R}$: the map $x\mapsto\tan\big(\pi(x-\tfrac12)\big)$ is a continuous bijection with a continuous inverse. So a bounded space can be homeomorphic to an unbounded one.
• $(0,1)\cong(a,b)$ for any $a<b$, by a linear map. Length, like boundedness, is invisible to the topology.
• The two-point Sierpiński space and the two-point discrete space are not homeomorphic: one has three open sets and the other four, and a homeomorphism matches the topologies one for one.

For a bijection the preimage of a set under the inverse map is the image of that set under the map itself. So asking the inverse to be continuous — preimages of open sets open — is asking images of open sets under the original map to be open, which is what being an open map means. The whole content is the identity between the image under one map and the preimage under the other, which holds only because the map is a bijection.

For a bijection, pulling back along the inverse is pushing forward along the map.

Write out what continuity of the inverse means, using the definition on the inverse map.

Substitute the identity from the first step.

That is precisely the definition of an open map, so the two conditions agree.

A homeomorphism is a bijection that is continuous and whose inverse is continuous. Both halves are needed.

• A homeomorphism is a bijection continuous in both directions; it matches the two topologies exactly.
• A continuous bijection is a homeomorphism exactly when it is an open map.
• $(0,1)\cong\mathbb{R}$: boundedness and length are not topological.
• Homeomorphic spaces have the same number of points and the same number of open sets.

Great work! You now know how to:

• define a homeomorphism and check both directions;
• give a continuous bijection that is not one;
• use the open-map criterion;
• show that boundedness and length are not preserved.