# Why is every map to an indiscrete space continuous?

Show that if $Y$ is a topological space, then every map $f:Y \rightarrow X$ is continuous when $X$ has the indiscrete topology.

Proof:

Assume $X$ has the indiscrete topology, $T=\{\varnothing,X\}$.

$f$ is continuous if $f^{-1}(V)$ is an open subset of $X$ whenever $V$ is an open subset of $Y$.

Let $V$ be an open subset of $Y$.

I dont know how to use this to show $f^{-1}(V)$ is an open subset of $X$.

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The definition of continuous is that $f\colon Y\to X$ then for every $U\subseteq X$ which is open, $f^{-1}(U)$ is open in $Y$. Now think which subsets of $X$ are open and what are their preimages? – Asaf Karagila Aug 14 '11 at 11:27
You may also want to accept some answers to previous questions, you can do that by clicking on the transparent check symbol next to the vote counter of an answer. Choose the answer most helpful to you, and if none is helpful enough please edit your question to indicate the missing parts in the answers so far. – Asaf Karagila Aug 14 '11 at 11:29

You got confused about the definition of continuity.

If $f\colon Y\to X$ is continuous then the preimage of open subsets of $X$ is open in $Y$.

Since $X$ has the indiscrete topology, we only have two open subsets. Namely, $X$ and $\varnothing$.

The preimage of the empty set is of course empty, and therefore open in $Y$. If we look at $f^{-1}(X) = \{y\in Y\mid f(y)\in X\}=Y$, and of course that $Y$ is open in $Y$.

Thus, $f$ is continuous regardless to the topology given on $Y$ whenever $X$ is indiscrete.

Exercise: Suppose $f\colon X\to Y$ and $X$ has the discrete topology, prove that $f$ is continuous.

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Thanks so much Asaf! – user8603 Aug 14 '11 at 11:47

To expand on Asaf's comment:

$f:X \rightarrow Y$ is continuous if $f^{-1}(O)$ is continuous for all open sets $O$ in $Y$. As $Y$ has the trivial topology, the only open sets are $\emptyset$ and $Y$. So to show that $f$ is continuous you need to show that $f^{-1}(\emptyset)$ and $f^{-1}(Y)$ are open, i.e. are in the topology of $X$.

A collection of sets is per definition a topology if it contains the entire space $X$ and $\emptyset$. $f^{-1}(\emptyset) = \emptyset$ and $f^{-1}(Y) = X$ are therefore both open and so $f$ is continuous.

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Just swap $X$ and $Y$ but names don't really matter... – Rudy the Reindeer Aug 14 '11 at 11:36
Thanks for your answers. Since I'm new to topology, how did you ascertain that $Y$ has the trivial topology? – user8603 Aug 14 '11 at 11:44
You were given that in your question, I swapped $X$ and $Y$. – Rudy the Reindeer Aug 14 '11 at 12:12