# Embeddings are precisely proper injective immersions.

We call a map $f: X \to Y$ between topological spaces proper if $f^{-1}(K)$ is compact for all compact $K \subset Y$. Where can I find a reference that embeddings are precisely proper injective immersions? Or could someone supply a proof?

• I'm not sure about this particular fact, but Bourbaki's volume on topology has a number of results on proper mappings. – user208259 Jan 23 '15 at 7:37
• This is not quite true, since, for example, the open unit disk is embedded in $\mathbb{R}^2$ by the inclusion, and the inverse image of the closed unit disk is the open unit disk. – user149792 Jan 23 '15 at 7:47

## 3 Answers

This is not quite true, since, for example, the open unit disk is embedded in $\mathbb{R}^2$ by the inclusion, and the inverse image of the closed unit disk is the open unit disk. What we will show is that embeddings with closed image are the same thing as proper injective immersions.

We note that the property of being closed in a topological space is a local conditions; that is, if $\{U_i\}_{i \in I}$ is an open cover of $X$, then $A$ is closed in $X$ if and only if $A \cap U_i$ is closed in $U_i$ for all $i$.

Similarly, the property of a bijection $f: X \to Y$ has the local property that if there is a set $I$ such that $\{X_i\}_{i \in I}$ and $\{Y_i\}_{i \in I}$ are open covers of $X$, $Y$ respectively indexed by $I$ with the property that $f|_{X_i}$ is a homeomorphism onto $Y_i$, then $f$ is a homeomorphism.

Now, suppose we have an embedding $f: X \to Z$ with closed image $Y$. By definition, an embedding is always an injective immersion, so we just need to show it is proper. Now, given any compact set $K$, $f^{-1}(K \cap Y)$ is compact since $K \cap Y$ is compact $($that is where we need that $Y$ is closed$)$ and $f^{-1}$ is a homeomorphism from $Y$ to $X$, so it takes compact sets to compact sets. Thus $f^{-1}(K) = f^{-1}(K \cap Y)$ is compact.

Conversely, suppose we have a proper injective immersion $X \to Z$, and suppose $Y$ is the image of $X$. Given any $P \in X$, we can find a neighborhood of $p$ such that $f|_U$ is an embedding, and shrinking $U$ if necessary, we can assume $U$ has compact closure contained in a single coordinate chart. Now, let $K$ be a compact subset of $Z$ whose interior contains $f(p)$. Then $f^{-1}(K) \setminus U$ is compact, so we can pick a finite open cover of $f^{-1}(K) \setminus U$ by open sets $f^{-1}(A_j)$, where each $A_j \subset Z$ is open but does not have $f(p)$ in its closure $($since $f$ is injective and $Z$ is Hausdorff, we can find such a cover$)$. It follows that there is some neighborhood $V$ of $f(p)$ that does not intersect any of the sets $A_j$, and we may assume $V$ is contained in the interior of $K$. It follows that $f^{-1}(V \cap K \cap Y) \subset f^{-1}(K) \setminus(f^{-1}(K) \cap U) = U$. Since $f|_U$ is a homeomorphism onto its image, $f|_{f^{-1}(V)}$ is also. The result then follows from the local characterization of homeomorphisms.

Closed embeddings (equivalently, embeddings with closed image) are precisely proper injective immersions.

$(\Rightarrow)$ If $f : M \to N$ is an embedding with $f(M)$ closed, then it is obviously proper (since the intersection of a compact set with $f(M)$ is compact).

$(\Leftarrow)$ On the other hand, if $f : M \to N$ is a proper, injective immersion, it is obviously closed (since proper maps are closed). We check that it's an embedding. (This is the only nontrivial part.)

Let $p \in M$; it suffices to find a neighborhood $U \subset M$ of $p$ and a neighborhood $V \subset N$ of $f(p)$ such that $V \cap M = f(U)$. We'll use appropriate coordinate charts and I will happily ignore the distinction between things going on in the manifolds and things going on "down in $\mathbb{R}^n$".

The fact that $f$ is an immersion guarantees that there are coordinates $x^1, \dots, x^m$ on a neighborhood $U$ of $p$ (with $p$ mapping to $0$), and coordinates $y^1, \dots, y^n$ on a neighborhood $V$ of $f(p)$ (with $f(p)$ mapping to $0$) such that, in these coordinates, the map $f$ takes the form $$(x^1, \dots, x^m) \mapsto (x^1, \dots, x^m, 0, \dots, 0),$$ and $f(U)$ is exactly the slice $S$ given by $y_{m+1} = \cdots = y_n = 0$. (This is a special case of the constant rank theorem.)

For $\epsilon > 0$ sufficiently small, the $\epsilon$-ball about $0$ in $\mathbb{R}^n$ is contained in $V$. We let $V_{\epsilon}$ denote this $\epsilon$-ball, and we let $U_\epsilon = f^{-1}(V_\epsilon \cap S)$, which is a neighborhood of $p$.

If, for some $\epsilon$, we have $V_{\epsilon} \cap f(M) = V_\epsilon \cap S$, then we are done. So assume otherwise; we will show this leads to a contradiction. For each large enough $n$ we can find $q_n \in V_{1/n} \cap f(M)$, say $q_n = f(p_n)$, such that $q_n \notin S$. Obviously $q_n \to f(p)$. Thus the set $K$ consisting of all of the $q_n$ together with $f(p)$ is compact in $N$.

However, $f^{-1}(K)$ is not compact, since no subsequence of the $p_n$ can converge; if there were such a subsequence, say $p_{n_k} \to p'$, then we would have (by continuity) $f(p') = f(p)$. Since $f$ is injective, $p' = p$. But by hypothesis none of the $f(p_{n_k})$ are in $S$, so none of the $p_{n_k}$ are in $U$, and hence they cannot converge to $p$.

As far as I concerned ,this property of embedding requires that $f(X)$ is closed in Y.

There is a counter-example .(When $f(X)$ is not closed in Y)

Let Y = $\{k_\lambda : \lambda \in A\}$ where A is a uncountable set. Let X be $\{k_1,\cdots,k_n,\cdots\}$ , an infinite countable subset of A.

Topology of X is discrete topology , hence X is not compact in X.

And topology of Y is defined as following:

$\{Y,\emptyset, \{\forall k_i \in X:\{k_i\}\}\}$ is basis of topology of Y. ($k_i$ is element of X) . Hence Y is a compact set in Y.

Define a map $f:X \rightarrow Y , f(x)=x$ . It's obvious that $f$ is an embedding.

However,$f^{-1} (Y) =X$ is not compact in X.