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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.

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    $\begingroup$ 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 $Y$. $\endgroup$
    – Bill Johnson
    Commented Jan 3, 2012 at 19:32
  • $\begingroup$ @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? $\endgroup$
    – Jacques Carette
    Commented Jan 3, 2012 at 20:15
  • $\begingroup$ 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 $f$. $\endgroup$
    – Bill Johnson
    Commented Jan 4, 2012 at 18:44
  • $\begingroup$ 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. $\endgroup$
    – Jacques Carette
    Commented Jan 4, 2012 at 20:12

1 Answer 1

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It's possible to find a sufficient condition without putting any restraint on the spaces $X$ or $Y$.

Let $Z$ be the Sierpinski space, that is $Z$ has $\{0,1\}$ as a base set, and $\tau = \{\emptyset, \{0\}, \{0,1\}\}$ as a topology.

If $f: X\to Y$ is not continuous, then there is $V\subseteq Y$ open such that $f^{-1}(V)$ is not open in $X$. Defining $g_V: Y\to Z$ as mapping $V$ to $0$ and $Y\setminus V$ to $1$ we immediately see that $g_V\circ f: X \to Z$ is not continuous.

An interesting (and probably solved) question would be how to characterise the spaces $Z$ that have this universal property for all spaces $X, Y$.

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