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I have seen the proof of the reverse here, but I cannot quite prove it the other way around.

I was thinking about using the fact that, for every $x \in X$, there is a ball $B_x$ where $f(B_x)=c\in \mathbb{R}^n$, and then I could argue that $X=\cup B_x$. Since every $B_x$ is connected and has more than one point from $X$, I would conclude that $X$ is the reunion of connected subsets, each one with at least one point in common with another one.

Is that approach correct? I feel like I have missed something around.

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AspiringMathematician
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    $f$ is defined only on $X$. So you'd need to look at $B_x \cap X$, but that need not be connected. Prove the contrapositive, if $X$ is not connected, there is a non-constant locally constant map. – Daniel Fischer Nov 16 '16 at 23:23

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To remove this trivial question from the unaswered list I remark that if $X$ is not connected then it is a union of two its non-empty disjoint open subspaces $X_1$ and $X_2$. Then a locally constant map $f:X\to\Bbb R$ such that $f(X_i)=i$ is not globally constant.

Alex Ravsky
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