Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$, $Y$, and $Z$ be topological spaces. Let $f:X \rightarrow Y$. Further assume that for every continuous function $g:Y \rightarrow Z$, $g \circ f$ is continuous.

Question: Under what conditions on the topology of $Y$ and/or $Z$ can we conclude that $f$ must then be continuous?

This is easily achieved if $Y$ is (essentially) homeomorphic with $Z$, but this seems like drastic overkill.

To me, this seems like some kind of 'universal property' for the continuity of $f$; or maybe some kind of generalized "uniformly continuous" condition? Note: original question due to Nick James, but he's not a big web user, so I am asking on his behalf.

share|improve this question
It is enough that $Y$ be homeomorphic to a subset of some product of $Z$, so a simple sufficient condition is that $Y$ be completely regular and $Z=[0,1]$. More generally, it is sufficient that there be enough continuous functions from $Y$ to $Z$ to separate points in $Y$ from disjoint closed subsets of $Y$. –  Bill Johnson Jan 3 '12 at 19:32
@Bill: thanks. But somehow that still seems like asking 'too much' for what you get. For example, why 'regular' but not Hausdorff or something weaker still? –  Jacques Carette Jan 3 '12 at 20:15
The second condition is sufficient with no assumptions on the topologies, Jacques. The first condition is just the observation that the second condition applies when $Y$ is completely regular and $Z=[0,1]$, which I (perhaps wrongly) assumed would take care of any situation that arose in Computer Science. $$ $$ I haven't thought about it, but would guess that the second condition is necessary in order for what you want to be true for all $X$ and all $f$. –  Bill Johnson Jan 4 '12 at 18:44
Thanks for the expansion. The situation where this arises concerns 'generalized Stream' spaces, $Y = \mathbb{T}\rightarrow Q$ where $\mathbb{T}$ is time-like (might be $\mathbb{R}$, or $\mathbb{R}^+$ but also much weirder spaces), and $Q$ is some kind of (topological) algebra. –  Jacques Carette Jan 4 '12 at 20:12
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.