Return to Question

Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$.

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

Motivation. Suppose you study a space that appears naturally, and you manage to bijectively parametrize with a continuous parameter that takes values is a nice parameter space (such as $l^2$). If you discover that the parametrization isn't a homeomorphism, you may wonder if you have proved anything at all.

Remarks (to the original question):

1. $Y$ is path-connected; in fact, no countable (or finite) subset can separate $Y$. (Here is why: the preimage of a countable subset of $Y$ is a countable subset of $X$. Then it is known that the complement in $l^2$ of a countable subset is homeomorphic to $l^2$, and hence the image of the complement in $Y$ is path-connected.)

2. $Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a result of Savkin that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent paper by Creswell in Monthly.

3. The same question is interesting when $X$ is $\mathbb R^n$. (I briefly thought that the answer in this case follows from the invariance of domain, and Bill kindly corrected me in comments).

Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$.

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

Motivation. Suppose you study a space that appears naturally, and you manage to bijectively parametrize with a continuous parameter that takes values is a nice parameter space (such as $l^2$). If you discover that the parametrization isn't a homeomorphism, you may wonder if you have proved anything at all.

Remarks (to the original question):

1. $Y$ is path-connected; in fact, no countable (or finite) subset can separate $Y$. (Here is why: the preimage of a countable subset of $Y$ is a countable subset of $X$. Then it is known that the complement in $l^2$ of a countable subset is homeomorphic to $l^2$, and hence the image of the complement in $Y$ is path-connected.)

2. $Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a result of Savkin [A criterion for Banach manifolds with a separable model to be finite-dimensional,Ukrainian Mathematical Journal volume 46, pages 1210-1214 (1994)] that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent paper by Creswell in Monthly.

3. The same question is interesting when $X$ is $\mathbb R^n$. (I briefly thought that the answer in this case follows from the invariance of domain, and Bill kindly corrected me in comments).

Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$.

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

Remarks:

1. $Y$ is path-connected.

2. $Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a result of Savkin that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent paper by Creswell in Monthly.

3. If $X$ is $\mathbb R^n$, then the bijection is a homeomorphism by invariance of domain.

Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$.

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

Motivation. Suppose you study a space that appears naturally, and you manage to bijectively parametrize with a continuous parameter that takes values is a nice parameter space (such as $l^2$). If you discover that the parametrization isn't a homeomorphism, you may wonder if you have proved anything at all.

Remarks  (to the original question):

1. $Y$ is path-connected; in fact, no countable (or finite) subset can separate $Y$. (Here is why: the preimage of a countable subset of $Y$ is a countable subset of $X$. Then it is known that the complement in $l^2$ of a countable subset is homeomorphic to $l^2$, and hence the image of the complement in $Y$ is path-connected.)

2. $Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a result of Savkin that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent paper by Creswell in Monthly.

3. The same question is interesting when $X$ is $\mathbb R^n$. (I briefly thought that the answer in this case follows from the invariance of domain, and Bill kindly corrected me in comments).

Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$.

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

Remarks:

1. $Y$ is path-connected.

2. $Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a result of Savkin that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent paper by Creswell in Monthly.

3. If $X$ is $\mathbb R^n$, then the bijection is a homeomorphism by invariance of domain.

Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$.

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

Remarks:

1. $Y$ is path-connected.

2. $Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a result of Savkin that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent paper by Creswell in Monthly.

3. If $X$ is $\mathbb R^n$, then the bijection is a homeomorphism by invariance of domain.