Section 5.6 (0405): Submersive maps

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:

it is the weakest topology on $Y$ such that $f$ is continuous,
the open subsets of $Y$ are $f^{-1}(U)$ for $U \subset X$ open,
the closed subsets of $Y$ are the sets $f^{-1}(Z)$ for $Z \subset X$ closed.

Proof. The set $\mathcal{T} = \{ f^{-1}(U) | U \subset X \text{ open}\} $ is a topology on $Y$. Firstly, $\emptyset = f^{-1}(\emptyset )$ and $f^{-1}(X) = Y$. So $\mathcal{T}$ contains $\emptyset $ and $Y$.

Now let $\{ V_ i\} _{i \in I}$ be a collection of open subsets where $V_ i \in \mathcal{T}$ and write $V_ i = f^{-1}(U_ i)$ where $U_ i$ is an open subset of $X$, then

\[ \bigcup \nolimits _{i\in I} V_ i = \bigcup \nolimits _{i\in I} f^{-1}(U_ i) = f^{-1}\left(\bigcup \nolimits _{i\in I} U_ i\right) \]

So $\bigcup _{i\in I} V_ i \in \mathcal{T}$ as $\bigcup _{i\in I} U_ i$ is open in $X$. Now let $V_1, V_2 \in \mathcal{T}$. We have $U_1, U_2$ open in $X$ such that $V_1 = f^{-1}(U_1)$ and $V_2 = f^{-1}(U_2)$. Then

\[ V_1 \cap V_2 = f^{-1}(U_1) \cap f^{-1}(U_2) = f^{-1}(U_1 \cap U_2) \]

So $V_1 \cap V_2 \in \mathcal{T}$ because $U_1 \cap U_2$ is open in $X$.

Any topology on $Y$ such that $f$ is continuous contains $\mathcal{T}$ according to the definition of a continuous map. Thus $\mathcal{T}$ is indeed the weakest topology on $Y$ such that $f$ is continuous. This proves that (1) and (2) are equivalent.

The equivalence of (2) and (3) follows from the equality $f^{-1}(X \setminus E) = Y \setminus f^{-1}(E)$ for all subsets $E \subset X$. $\square$

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:

it is the strongest topology on $Y$ such that $f$ is continuous,
a subset $V$ of $Y$ is open if and only if $f^{-1}(V)$ is open,
a subset $Z$ of $Y$ is closed if and only if $f^{-1}(Z)$ is closed.

Proof. The set $\mathcal{T} = \{ V\subset Y | f^{-1}(V) \text{ is open}\} $ is a topology on $Y$. Firstly $\emptyset = f^{-1}(\emptyset )$ and $f^{-1}(Y) = X$. So $\mathcal{T}$ contains $\emptyset $ and $Y$.

Let $(V_ i)_{i \in I}$ be a family of elements $V_ i \in \mathcal{T}$. Then

\[ \bigcup \nolimits _{i\in I} f^{-1}(V_ i) = f^{-1}\left(\bigcup \nolimits _{i\in I} V_ i\right) \]

Thus $\bigcup _{i\in I} V_ i \in \mathcal{T}$ as $\bigcup _{i\in I}f^{-1}(V_ i)$ is open in $X$. Furthermore if $V_1, V_2 \in \mathcal{T}$ then

\[ f^{-1}(V_1) \cap f^{-1}(V_2) = f^{-1}(V_1 \cap V_2) \]

So $V_1 \cap V_2 \in \mathcal{T}$ because $f^{-1}(V_1) \cap f^{-1}(V_2)$ is open in $X$.

Finally a topology on $Y$ such that $f$ is continuous is included in $\mathcal{T}$ according to the definition of a continuous function, so $\mathcal{T}$ is the strongest topology on $Y$ such that $f$ is continuous. It proves that (1) and (2) are equivalent.

Finally, (2) and (3) equivalence follows from $f^{-1}(X\setminus E) = Y \setminus f^{-1}(E)$ for all subsets $E \subset X$. $\square$

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.

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).
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

$f^{-1}(\overline{T}) = \overline{f^{-1}(T)}$,
$T \subset Y$ is closed if and only if $f^{-1}(T)$ is closed,
$T \subset Y$ is open if and only if $f^{-1}(T)$ is open, and
$T \subset Y$ 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

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

Comments (2)

Comment #1328 by Hua WANG on February 24, 2015 at 17:52

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, $Y=\mathbb{R}, y_0=1, T=\{x\in\mathbb{R}, 0<x<1\}$ ), 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 Johan on February 25, 2015 at 12:53

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

The Stacks project

5.6 Submersive maps

Comments (2)

Post a comment