# 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 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
I know this is an old answer, but rather than making a completely new question I thought I'd ask here instead. How do we know that $f^{-1}(X) = Y$? – Irregular User May 19 at 8:38
Because for every $y\in Y$, $f(y)\in X$. – Asaf Karagila May 19 at 8:39

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