# (Proof-verification) Constant maps are the only maps that are always continuous

Given two spaces $X,Y$, I know that, regardless of the choice of topologies $\tau_X$, $\tau_Y$ for $X$ and $Y$, one has that the constant maps are continuous.

What I want to show is that this is the only class of maps for which this property holds.

Claim: Let the topology $\tau_X$ on $X$ be fixed. If a map $X \to Y$ is continuous when $Y$ is given the discrete topology, then it is continuous for any choice of topology on $Y$.
Proof of Claim: If $U \subset Y$ is open for any given choice of topology on $Y$, it is open in the discrete topology on $Y$, since the discrete topology is at least as fine as all other topologies. Thus if the inverse images of open sets in the discrete topology are open in $X$, then the inverse images of open sets of any other topology $\tau_Y$ are open in $X$.

Claim: Let the topology $\tau_Y$ be fixed. If a map $X \to Y$ is continuous when $X$ is given the trivial topology, then it is continuous when $X$ is given any other topology.
Proof of claim: This means that the inverse images of all open sets in $\tau_Y$ are either $\emptyset \subset X$ or $X \subset X$. Since the empty set and $X$ itself are contained in any topology on $X$, it follows that the map is continuous given any choice of topology for $X$.

Corollary: If a map $X \to Y$ is continuous when $Y$ is given the discrete topology and $X$ is given the trivial topology, then it is continuous for any choice of topologies on $X$ and $Y$.

Claim: If a map $f: X \to Y$ is continuous when $X$ has the trivial topology and $Y$ has the discrete topology, then $f$ is a constant map.
Proof of Claim: The inverse image of the empty set is always the empty set. Having dispensed of that trivial case, any other open set in $Y$ is the arbitrary union of singleton sets of $Y$. If the inverse image of any point is open, then the inverse image of any open set is open since $$f^{-1}(Y)=f^{-1}\left(\bigcup \{y\}\right)=\bigcup f^{-1}(\{y\})$$ which is the arbitrary union of open sets in $X$ hence open (even if $X$ didn't have the trivial topology). Thus a map into a space with the discrete topology is continuous if and only if the inverse image of every singleton set is an open set. In this case it means that $f$ is continuous if and only if the inverse image of every singleton is either empty or the entire space $X$. Assume by means of contradiction that $f$ were not constant, i.e. mapped to at least two distinct points $y_1 \not= y_2$. Then $f^{-1}(y_1)\not=\emptyset, f^{-1}(y_1)\not=X \supset f^{-1}(y_1)\cup f^{-1}(y_2) \supsetneq f^{-1}(y_1)$ which is a contradiction because $f$ is supposed to be continuous. Therefore $f$ is constant. $\square$

• It looks good to me – shai horowitz Oct 31 '16 at 22:31
• You've shown that if $f:X\to Y$ is continuous when $X$ has the indiscrete topology and $Y$ the discrete topology, then $f$ is continuous for arbitrary topologies on $X$ and $Y$ and is constant; you've not shown that the constant functions are the only functions continuous for all topologies on $X$ and $Y$. For that you should show that if $f$ is not constant, there is a choice of topologies making it not continuous. – Brian M. Scott Oct 31 '16 at 22:31

It seems that you want to prove that the only functions continuous from $X$ to $Y$ in all the topologies are the constant ones. It is obvious that the constant functions are continuous. Let us show that they are the only ones.

Choose then the trivial topology on $X$ (i.e. $\tau _X = \{ \emptyset, X\}$) and the discrete topology on $Y$ (i.e. the topology of all the subsets of $Y$). Let $f$ be a continuous function.

Clearly, you cannot have $f^{-1} (\{y\}) = \emptyset$ for all $y \in Y$, because $f \ne \emptyset$, so there exist $y \in Y$ with $f^{-1} (\{y\}) \ne \emptyset$. By our choice of $\tau _Y$, $\{y\}$ is open and since $f$ is continuous, then $f^{-1} (\{y\})$ must be open, so the only possibility is $f^{-1} (\{y\}) = X$, which means that $f$ is constant (equal to $y$).