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How can I prove that if $f: X \to Y$ is locally constant and $X$ is connected then $f$ is constant?

Definition of locally constant: for all $x_0$ in $X$ there is a neighborhood $U$ of $x_0$ then for all $x\in{U}$ $f(x)=f(x_0)$.

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  • $\begingroup$ Consider $A=\{x\in X\mid f(x)=f(x_0)\}$. Is it open? Is it closed? $\endgroup$ – Thomas Shelby Jan 4 at 17:11
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Let’s prove that a locally constant map is constant on connected components.

Take a connected component $C \subseteq X$ and $a \in C$. Denote $b = f(a)$ and let’s prove that $f(x)=b$ for $x \in C$. By hypothesis, $U=\{x \in C \mid f(x)=b\}$ is open in $C$. Now consider $V=\cup_{Y \setminus \{b\}} V_y$ where $V_y$ is an open subset of $C$ where $f$ is constant with value equal to $y$. $V$ is an open subset of $C$. $V$ is empty. If it was not, $U,V$ would be two non empty open subsets of $C$ with empty intersection. A contradiction with the fact that $C$ is a connected component.

$f$ is constant on all connected components of $C$ which is supposed to be connected. Hence $f$ is constant.

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