[proofplan] We prove the contrapositive by assuming that $f$ is not constant. Then two points of $X$ have distinct images in $Y$, and the $T_0$ property separates these two image points by an [open set](/page/Open%20Set) containing exactly one of them. Continuity pulls this open set back to an open subset of the indiscrete space $X$, but the preimage is simultaneously nonempty and proper, contradicting the fact that the only open subsets of $X$ are $\varnothing$ and $X$. [/proofplan] [step:Assume two domain points have distinct images] Suppose, for contradiction, that $f$ is not constant. Since $f:X\to Y$ is a map and $X$ is nonempty, there exist points $x_1,x_2\in X$ such that $f(x_1)\neq f(x_2)$. [guided] To prove that $f$ is constant, we must show that every pair of points in $X$ has the same image. We argue by contradiction. Assume that $f$ is not constant. By the definition of a nonconstant map, there must be two points $x_1,x_2\in X$ such that their images are distinct: \begin{align*} f(x_1)\neq f(x_2). \end{align*} The hypothesis that $X$ is nonempty ensures that the domain is not vacuous, but the contradiction argument itself starts from the stronger assumption that such two points exist. [/guided] [/step] [step:Use the $T_0$ property to separate the two image points] Because $(Y,\tau_Y)$ is $T_0$ and $f(x_1)\neq f(x_2)$, there exists an open set $U\in\tau_Y$ such that either $f(x_1)\in U$ and $f(x_2)\notin U$, or $f(x_2)\in U$ and $f(x_1)\notin U$. [/step] [step:Pull the separating open set back to a nonempty proper subset of $X$] Define the subset $A\subset X$ by \begin{align*} A:=f^{-1}(U)=\{x\in X:f(x)\in U\}. \end{align*} Since $U\in\tau_Y$ and $f:X\to Y$ is continuous, the preimage $A=f^{-1}(U)$ is open in the topology $\{\varnothing,X\}$ on $X$. If $f(x_1)\in U$ and $f(x_2)\notin U$, then $x_1\in A$ and $x_2\notin A$. If $f(x_2)\in U$ and $f(x_1)\notin U$, then $x_2\in A$ and $x_1\notin A$. In either case, $A$ is nonempty and $A\neq X$. [guided] The open set $U$ lives in the codomain $Y$, so we convert it into a subset of the domain by taking its preimage. Define \begin{align*} A:=f^{-1}(U)=\{x\in X:f(x)\in U\}. \end{align*} Continuity of $f:X\to Y$ means precisely that the preimage of every open set in $Y$ is open in $X$. Since $U\in\tau_Y$, it follows that $A$ is open in $X$. Now we record what the separation property tells us about $A$. If $f(x_1)\in U$ and $f(x_2)\notin U$, then by the definition of preimage we have $x_1\in A$ and $x_2\notin A$. Hence $A$ is nonempty, because it contains $x_1$, and $A$ is not all of $X$, because it does not contain $x_2$. The other possible $T_0$ separation is symmetric: if $f(x_2)\in U$ and $f(x_1)\notin U$, then $x_2\in A$ and $x_1\notin A$. Again $A$ is nonempty and proper. Thus in every case, \begin{align*} \varnothing\subsetneq A\subsetneq X. \end{align*} [/guided] [/step] [step:Contradict the indiscrete topology and conclude constancy] The topology on $X$ is the [indiscrete topology](/page/Indiscrete%20Topology) $\{\varnothing,X\}$. Therefore every open subset of $X$ is either $\varnothing$ or $X$. This contradicts the previous step, where $A=f^{-1}(U)$ was shown to be open, nonempty, and not equal to $X$. Hence the assumption that $f$ is not constant is false. Therefore, for all $x_1,x_2\in X$, one has $f(x_1)=f(x_2)$, so $f$ is constant. [/step]