# Embeddings into a Hausdorff space are proper?

I am sure I am missing something obvious here, but I cannot resolve the following (presumably apparent) contradiction. In the book "Topological manifold" by John Lee, 2nd edition, proposition 4.93, some sufficient conditions for a map to be proper are stated. Condition (e) is

Let $F:X\rightarrow Y$ be a continuous map, $Y$ a Hausdorff space. If $F$ has a continuous left inverse than $F$ is proper.

It seems to me that since $F$ is continuous with left continuous inverse it is a topological embedding. But then wouldn't the example mentioned here

of the embedding of an open disk (with subspace topology) in $\mathbb{R}^2$ be a counterexample as the inclusion is not proper? (e.g. the inverse image of the closed disk is the open disk).

In fact condition (d) of proposition 4.93 says that closed embeddings are proper.

• What is the supposed continuous left inverse $g\colon \mathbb R^2\to D$ such that $g\circ f=\operatorname{id}_D$? – Hagen von Eitzen Aug 5 '15 at 16:07
• @HagenvonEitzen Your comment make me think I was wrong in assuming the inclusion $f$ to be a topological embedding. However in order for $f$ to be a topological embedding isn't enough that $U\subset D$ is open if and only if $f(U)$ is open in $\mathbb{R}^2$, which holds $D$ being open in $\mathbb{R}^2$? – GFR Aug 5 '15 at 16:21
• @GFR: You're ignoring one of the assumptions, which is that $F$ has a left inverse. $F$ being an embedding is not enough to guarantee the existence of a left inverse, which is what Hagen is pointing out. – user98602 Aug 5 '15 at 17:01
• Thanks, my confusion was indeed because I did not realise that an embedding does not necessarily have a left inverse. – GFR Aug 5 '15 at 18:12

I think you're confusing some implications with their converses. My condition (e) says if $F$ has a continuous left inverse, then $F$ is proper. It's also true in that case that $F$ is a topological embedding (see Problem 3-13 in the same book). However, neither of those implications is reversible. There exist nonproper embeddings that have no continuous left inverse (such as the embedding of an open disk into $\mathbb R^2$ that you mentioned), and there exist proper embeddings that have no continuous left inverse (such as the embedding of the unit circle into $\mathbb R^2$). There's no contradiction.