
## 5.6 Submersive maps

If $X$ is a topological space and $E \subset X$ is a subset, then we usually endow $E$ with the induced topology.

Lemma 5.6.1. Let $X$ be a topological space. Let $Y$ be a set and let $f : Y \to X$ be an injective map of sets. The induced topology on $Y$ is the topology characterized by each of the following statements:

1. it is the weakest topology on $Y$ such that $f$ is continuous,

2. the open subsets of $Y$ are $f^{-1}(U)$ for $U \subset X$ open,

3. the closed subsets of $Y$ are the sets $f^{-1}(Z)$ for $Z \subset X$ closed.

Proof. Omitted. $\square$

Dually, if $X$ is a topological space and $X \to Y$ is a surjection of sets, then $Y$ can be endowed with the quotient topology.

Lemma 5.6.2. Let $X$ be a topological space. Let $Y$ be a set and let $f : X \to Y$ be a surjective map of sets. The quotient topology on $Y$ is the topology characterized by each of the following statements:

1. it is the strongest topology on $Y$ such that $f$ is continuous,

2. a subset $V$ of $Y$ is open if and only if $f^{-1}(V)$ is open,

3. a subset $Z$ of $Y$ is closed if and only if $f^{-1}(Z)$ is closed.

Proof. Omitted. $\square$

Let $f : X \to Y$ be a continuous map of topological spaces. In this case we obtain a factorization $X \to f(X) \to Y$ of maps of sets. We can endow $f(X)$ with the quotient topology coming from the surjection $X \to f(X)$ or with the induced topology coming from the injection $f(X) \to Y$. The map

$(f(X), \text{quotient topology}) \longrightarrow (f(X), \text{induced topology})$

is continuous.

Definition 5.6.3. Let $f : X \to Y$ be a continuous map of topological spaces.

1. We say $f$ is a strict map of topological spaces if the induced topology and the quotient topology on $f(X)$ agree (see discussion above).

2. We say $f$ is submersive1 if $f$ is surjective and strict.

Thus a continuous map $f : X \to Y$ is submersive if $f$ is a surjection and for any $T \subset Y$ we have $T$ is open or closed if and only if $f^{-1}(T)$ is so. In other words, $Y$ has the quotient topology relative to the surjection $X \to Y$.

Lemma 5.6.4. Let $f : X \to Y$ be surjective, open, continuous map of topological spaces. Let $T \subset Y$ be a subset. Then

1. $f^{-1}(\overline{T}) = \overline{f^{-1}(T)}$,

2. $T \subset Y$ is closed if and only if $f^{-1}(T)$ is closed,

3. $T \subset Y$ is open if and only if $f^{-1}(T)$ is open, and

4. $T \subset Y$ is locally closed if and only if $f^{-1}(T)$ is locally closed.

In particular we see that $f$ is submersive.

Proof. It is clear that $\overline{f^{-1}(T)} \subset f^{-1}(\overline{T})$. If $x \in X$, and $x \not\in \overline{f^{-1}(T)}$, then there exists an open neighbourhood $x \in U \subset X$ with $U \cap f^{-1}(T) = \emptyset$. Since $f$ is open we see that $f(U)$ is an open neighbourhood of $f(x)$ not meeting $T$. Hence $x \not\in f^{-1}(\overline{T})$. This proves (1). Part (2) is an easy consequence of (1). Part (3) is obvious from the fact that $f$ is open and surjective. For (4), if $f^{-1}(T)$ is locally closed, then $f^{-1}(T) \subset \overline{f^{-1}(T)} = f^{-1}(\overline{T})$ is open, and hence by (3) applied to the map $f^{-1}(\overline{T}) \to \overline{T}$ we see that $T$ is open in $\overline{T}$, i.e., $T$ is locally closed. $\square$

Lemma 5.6.5. Let $f : X \to Y$ be surjective, closed, continuous map of topological spaces. Let $T \subset Y$ be a subset. Then

1. $\overline{T} = f(\overline{f^{-1}(T)})$,

2. $T \subset Y$ is closed if and only if $f^{-1}(T)$ is closed,

3. $T \subset Y$ is open if and only if $f^{-1}(T)$ is open, and

4. $T \subset Y$ is locally closed if and only if $f^{-1}(T)$ is locally closed.

In particular we see that $f$ is submersive.

Proof. It is clear that $\overline{f^{-1}(T)} \subset f^{-1}(\overline{T})$. Then $T \subset f(\overline{f^{-1}(T)}) \subset \overline{T}$ is a closed subset, hence we get (1). Part (2) is obvious from the fact that $f$ is closed and surjective. Part (3) follows from (2) applied to the complement of $T$. For (4), if $f^{-1}(T)$ is locally closed, then $f^{-1}(T) \subset \overline{f^{-1}(T)}$ is open. Since the map $\overline{f^{-1}(T)} \to \overline{T}$ is surjective by (1) we can apply part (3) to the map $\overline{f^{-1}(T)} \to \overline{T}$ induced by $f$ to conclude that $T$ is open in $\overline{T}$, i.e., $T$ is locally closed. $\square$

[1] This is very different from the notion of a submersion between differential manifolds! It is probably a good idea to use “strict and surjective” in stead of “submersive”.

Comment #1328 by Hua WANG on

Lemma 5.5.5 (1) does not hold. Here's a counter-example. Suppose $Y$ is a topological space which admits a closed point $y_0\in Y$ such that $y_0\in \overline{T}\backslash T$ for a $T\subset Y$(e.g, $Y=\mathbb{R}, y_0=1, T=\{x\in\mathbb{R}, 0), and $X$ the disjoint sum of $Y$ and a point $\ast$. Let $f:X\rightarrow Y$ be defined as $f(\ast)=y_0$ and $f(y) = y,\forall y\in Y$. Then $f$ is closed and surjective, but $f^{-1}(T)=T$ so $\overline{f^{-1}(T)}=\overline{T}$, while $f^{-1}(\overline{T})=\overline{T}\cup\{\ast\}$.

However, (4) in that lemma remains true, for to make the arguments there work, we only need the restriction of $f$, or more precisely the map induced by $f$, $\overline{f^{-1}(T)}\rightarrow \overline{T}$, is still submersive as defined above, but this will hold so long as $f(\overline{f^{-1}(T)}) = \overline{T}$, which is the arguments in (1) really show.

Comment #1329 by on

Good catch! I fixed it here following your suggestion. Thanks very much.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).