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Let $X$, and $Y$ be topological spaces, and let $f:X\rightarrow Y$ be a continuous and one-to-one map.

When is $X$ homeomorphic to $f(X)$?

Stefan Hamcke
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Ronald
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    Sufficient criteria are easy - or criteria for $f$ to be a homeomorphism. But if $f$ is not a homeomorphism, a different map might still be. – Daniel Fischer Aug 04 '13 at 21:23
  • An interesting base case is when $Y=X$, as sets but $Y$ has a different topology, and $f=\mathrm{id}_X$. – Thomas Andrews Aug 04 '13 at 21:35
  • Because of the properties of the subspace topology, it suffices to consider the case where $f$ is a bijection. – kahen Aug 04 '13 at 21:36
  • As an example, let $X = Y = \mathbb{R}^\infty$ the space of all real sequences with only finitely many nonzero terms, endowed with the topology induced by the norm $\lVert x\rVert = \sqrt{\sum x_k^2}$, and $f(x) = \left( \frac{x_k}{k+1}\right)_{k\in\mathbb{N}}$. Then $f$ is continuous and bijective, but not a homeomorphism. But the identity of $X$ obviously is a homeomorphism. – Daniel Fischer Aug 04 '13 at 21:37

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Well... when $f$ is open (or closed). A nice criterion is: $X$ compact, $Y$ Hausdorff, then $f$ is a closed map. Indeed let $C\subset X$ be closed, then $C$ is compact. The continuous image of a compact set is compact, so $f(C)\subset Y$ is compact, and thus closed.

Note: I interpreted your question as: "When is $f$ an homeomorphism with its image?" Obviously as Daniel Fisher stated in a comment, $X$ and $f(X)$ can be homeomorphic without $f$ being a homeomorphism.

Daniel Robert-Nicoud
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