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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 $X$.

user8603
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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

2 Answers2

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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.

Asaf Karagila
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  • 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 '16 at 08:38
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    Because for every $y\in Y$, $f(y)\in X$. – Asaf Karagila May 19 '16 at 08:39
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    I guess we're assuming that $f$ is defined for all $y \in Y$. Suppose $f$ is only defined on non-open set $C$, then $f^{-1}(X) = C$, and $f$ is not continuous. – Edward Newell Feb 18 '21 at 20:03
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    @EdwardNewell: When we write $f\colon A\to B$, the notation reads that the domain of $f$ is all of $A$. So yes, if $f\colon Y\to X$ is continuous, then it is defined on all of $Y$. – Asaf Karagila Feb 18 '21 at 20:09
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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.

Rudy the Reindeer
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