When Is a Proper Injective Map a Homeomorphism onto Its Image?

Let $f:X\to Y$ be continuous, injective, and proper, where proper means that the inverse image of every compact subset of $Y$ is compact. The natural question is whether $f$ is a homeomorphism from $X$ onto its image $f(X)$, equipped with the subspace topology.

The answer is yes if $Y$ is Hausdorff and first-countable. The Hausdorff assumption is essential: first countability alone does not suffice.

The corrected theorem

Assume that:

• $f:X\to Y$ is continuous;
• $f$ is injective;
• $f$ is proper;
• $Y$ is Hausdorff and first-countable.

Then

$$f:X\longrightarrow f(X)$$

is a homeomorphism, where $f(X)$ has the subspace topology inherited from $Y$.

Because $f$ is already continuous and injective, only continuity of the inverse

$$f^{-1}:f(X)\to X$$

needs to be proved.

Why first countability helps

A map from a first-countable space is continuous if it preserves convergent sequences. More precisely, since $f(X)$ is a subspace of the first-countable space $Y$, it is also first-countable. Therefore it is enough to show:

Whenever $y_n\to y$ in $f(X)$, the corresponding sequence $f^{-1}(y_n)$ converges to $f^{-1}(y)$ in $X$.

Set

$$x=f^{-1}(y),\qquad x_n=f^{-1}(y_n).$$

The proof will use properness and compactness.

The compact convergent-sequence argument

Define

$$K=\{y\}\cup\{y_n:n\in\mathbb N\}.$$

The set $K$ is compact. Indeed, if an open cover contains a set covering $y$, convergence $y_n\to y$ implies that this one open set covers all but finitely many $y_n$. The remaining finitely many points can be covered by finitely many members of the cover.

Since $f$ is proper,

$$C=f^{-1}(K)$$

is compact in $X$.

The restriction

$$f|_C:C\to K$$

is continuous and bijective. It is bijective because $f$ is injective and because every point of $K$ belongs to $f(X)$. Since $C$ is compact and $K$ is Hausdorff, the standard compact-to-Hausdorff theorem says that $f|_C$ is a homeomorphism.

Consequently, its inverse is continuous, and therefore

$$x_n=(f|_C)^{-1}(y_n)\longrightarrow (f|_C)^{-1}(y)=x.$$

Thus $f^{-1}$ preserves convergent sequences. Since $f(X)$ is first-countable, $f^{-1}$ is continuous. Hence $f$ is a homeomorphism onto its image.

A fully worked example

Consider

$$f:\mathbb R\to\mathbb R^2, \qquad f(t)=(t,t^2).$$

This map is continuous and injective. It is also proper. To see this, let $K\subseteq\mathbb R^2$ be compact. Its first-coordinate projection is compact and therefore bounded, so there is an $M>0$ such that every $(x,y)\in K$ satisfies $|x|\le M$. If $f(t)\in K$, then $|t|\le M$, and hence

$$f^{-1}(K)\subseteq[-M,M].$$

The set $f^{-1}(K)$ is closed in the compact interval $[-M,M]$, because $K$ is closed and $f$ is continuous. Therefore $f^{-1}(K)$ is compact.

The theorem now shows that

$$t\longmapsto(t,t^2)$$

is a homeomorphism from $\mathbb R$ onto the parabola

$$\{(x,x^2):x\in\mathbb R\}$$

with its subspace topology.

Why Hausdorffness cannot be omitted

First countability by itself is not enough. Let

$$X=\{a,b\}$$

with the discrete topology, and let

$$Y=\{0,1\}$$

with the indiscrete topology $\{\varnothing,Y\}$. Define a bijection by

$$f(a)=0,\qquad f(b)=1.$$

The space $Y$ is finite, so it is compact and first-countable. The map $f$ is continuous because the only open subsets of $Y$ are $\varnothing$ and $Y$. It is proper: every subset of the finite space $Y$ is compact, and its inverse image is a finite subset of the discrete space $X$, hence compact.

However, $f$ is not a homeomorphism. Its inverse is not continuous, because

$$(f^{-1})^{-1}(\{a\})=\{0\},$$

and $\{0\}$ is not open in the indiscrete topology on $Y$.

Thus the original claim is false for arbitrary first-countable target spaces, but it becomes true after adding the Hausdorff condition.

Common mistakes

Assuming compact sets are enough without Hausdorffness

A continuous bijection from a compact space to a non-Hausdorff space need not be a homeomorphism. The compact-to-Hausdorff theorem specifically requires the target to be Hausdorff.

Applying properness to arbitrary subsets

Properness gives compactness of $f^{-1}(K)$ only when $K$ is compact. In the proof, the relevant compact set is the convergent sequence together with its limit:

$$K=\{y\}\cup\{y_n:n\in\mathbb N\}.$$

Forgetting the subspace topology

The conclusion is that $X$ is homeomorphic to $f(X)$ with the topology inherited from $Y$. It does not claim that $f(X)$ is open or closed in $Y$, unless additional assumptions imply that.

Confusing first countability with sequentiality

First countability is used to conclude that sequential continuity implies continuity. Without a condition of this kind, proving that the inverse preserves sequences may not be enough.