Locally constant map of topological spaces

Question

Question

A map of topological spaces $f: X \rightarrow Y$ is called locally constant if for every $x \in X$ there is an open neighbourhood $U_{x} \subset X$, x $\in U_{x}$, such that $f$ is constant on $U_{x}$. Prove that if $X$ is connected, then any locally constant map $f: X \rightarrow Y$ is constant, i.e. $f(X)$ is a point.

Proof.

Assume f is locally not globally constant. And let's restrict f to be surjective and it's image Y to have the discrete topology (this doesn't change the map itself). We can notice that f is continuous, since $$B = \{ \{y\} |y \in Y\}$$ is a basis for the topology of Y it's enough to show that $f^{-1}(\{y\})$ is open. But this is clear, since for any $x \in f^{-1}(\{y\})$ we have open $U_x \subset f^{-1}(\{y\})$. Thus f is continuous. Now let $$g: Y \rightarrow \{0,1\}$$ be a map, for which g(y') = 0 for some $y' \in Y$ and g(y) = 1 for $y \in Y$ and $y \neq y'$. Clearly map g is continuous (since all maps from discrete topological spaces are continuous). So $$g \circ f : X \rightarrow \{0,1\}$$ is a continuous, surjective map (surjective since we assumed f is not globally constant). This is a contradiction, since X is connected.

Doubts

Is this proof correct? I'm not sure I can change topologies just like that. The change doesn't affect the map and if a non-globally constant function f exists for any topology on y, then it also exists for the discrete one. I know there's a much simpler solution to this.

Answer 1

As you note the preimages of singletons under $f$ are open (because $f^{-1}(\{y\})=\bigcup\limits_{x\in f^{-1}(\{y\})}U_x$). From this you can almost directly get your conclusion because it implies that the preimage of any subset of $Y$ is open. Hence, if $y, y' \in f(X)$ with $y\neq y'$, $f^{-1}(\{y\})\cup f^{-1}(Y\backslash \{y\})$ is a seperation of $X$, contradicting connectedness. I think your proof is ok, but I think you are taking a detour, and making your proof longer than it has to be, when you start by assuming $Y$ is a discrete topoligcal space.

Answer 2

Your can show that $f$ is continuous by showing that the inverse image of any singleton is open (as you did) but you cannot assume that $Y$ has the discrete topology.

Beyond that, your solution is correct (perhaps you forgot to mention that you assume $f$ is not constant and therefore $y'$ is in the image of $f$ and there is also at least one element of $Y\setminus \{y'\}$ in the image)

Answer 3

The topology on $Y$ is completely irrelevant. In fact, we can prove the following:

Let $f : X \to Y$ be a locally constant function from a connected topological space $X$ to a set $Y$. Then $f$ is constant.

Let $x_0 \in X$, $U = \{x \in X \mid f(x) = f(x_0) \}$, $V = \{x \in X \mid f(x) \ne f(x_0) \}$. Then $U$ and $V$ are open in $X$ due to local constantness, $U \cap V = \emptyset$, $U \cup V = X$. Since $X$ is connected and $U \ne \emptyset$, we conclude $U = X$.

Answer 4

If you want to use the mappping characterisation of connectedness (with maps to discrete $\{0,1\}$), I'd have a different write-up:

Let $\mathcal{U}=\{U_x: x \in X\}$ be the open cover of $X$ of open sets such that $f\restriction_{U_x}$ is constant for each $U \in \mathcal{U}$. This is just a restatement of the fact that $f$ is locally constant. Suppose $f$ is not constant, then it has at least two values $c:=f(x_0) \neq f(x_1)$ for $x_0,x_1 \in X$.

Now define $g: X \to \{0,1\}$ directly by defining it separately on each $U_x \in \mathcal{U}$: $g(x)=0$ iff $f(x)=c$ and $g(x)=1$ iff $f(x) \neq c$. This $g$ is constant on each $U_x$ (as $f$ is constant there) and so $g\restriction_{U_x}$ is always continuous. The pasting lemma (for open covers; there is also a version for locally finite closed covers) then implies that $g$ is continuous on $X$ and non-constant (as witnessed by $f\{0,1\} = [\{x_0,x_1\}]\subseteq f[X]$) contradicting the connectedness of $X$.