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This is the problem: I have a space $X=\{a,b\}$ with trivial topology (open sets are empty set and whole $X$) and continuous map $f\colon X\to Y$. ($Y$ is arbitrary space). Can I conclude that $f(a)=f(b)$ because original of some open set in $Y$ is $\{a,b\}$, and original of some closed set in $Y$ is $\{a,b\}$? My friend used it but I don't agree.

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Generally you cannot use that argument. For instance we might ask $Y=X$ and we also give $Y$ the trivial topology. Then the identity map is continuous but $f(a)\neq f(b)$.

In general, when the target space is in trivial topology, all functions are continuous.

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No you cannot conclude this.

Consider the topology on $Y=\{a,b,c\}$ where $\{\varnothing,\{c\},\{a,b\},Y\}$ are open; and $f\colon X\to Y$ is the identity function.

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You can conclude that $f(a)$ and $f(b)$ are topologically indistinguishable, i.e., every open set either contains both $f(a)$ and $f(b)$ or contains neither. In general, you cannot conclude $f(a) = f(b)$. However, if $Y$ satisfies the $\mathrm{T}_0$ separation axiom, then $f(a) = f(b)$.

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