Is a function from a product space $X \times Y \to Z$ continuous if and only if precomposition with all continuous functions $X\to X\times Y$ and $Y\to X\times Y$ are continuous? Clearly one direction holds for all spaces and I think I've shown that the other direction is true for functions $f:\mathbb{R}^n \to \mathbb{R}^m$ but I don't know if this is true generally or just true for nice enough spaces. Any counterexamples or special cases where this is true would be greatly appreciated.

Update: I have shown this is true if your base spaces are path connected or locally path connected metric spaces using the Tietze extension theorem but I still don't know if this is true generally.

Update: In retrospect, this argument only really works for spaces where there are short enough paths between close points, such as a length space.

Suppose $X$ and $Y$ are path connected metric spaces. In metric spaces continuity is the same as sequential continuity. Take a convergent sequence $\{x_n\}$ and its limit $x_\infty$ in $X$ (or $Y$) and define a continuous map from $A=\{x_n\}\cup x_\infty$ to $\mathbb{R}$ which sends $x_n$ to $1/n$ and $x_\infty$ to 0. $A$ is a closed subset of $X$ so by the Tietze extension theorem it extends to a map $X\to \mathbb{R}$. Then given a convergent sequence $\{z_n\}$ which converges to $z_\infty$ in $X\times Y$ we can construct a map from $\{ \frac{1}{n}\} \cup 0$ to our sequence which sends $\frac{1}{n}$ to $z_n$ and extend it to a map from $[0, 1]$ to $X\times Y$ by connecting the paths between $z_n$ and $z_{n+1}$. We then extend that function to the entirety of $\mathbb{R}$, which when composed with our function $X \to \mathbb{R}$ gives us a continuous function which takes a convergent sequence in $X$ to a convergent sequence in $X\times Y$. If the image of the convergent sequence in $X$ under this map is convergent in $Z$, then the image of the convergent sequence in $X\times Y$ will also be convergent in $Z$.

  • $\begingroup$ it is not clear what exactly you are pre-composing? Is it the inclusion map? OR do you want the statement to hold for all continuous functions $X\to\ X\times Y$, as mentioned in your post? Later seems very strong requirement! $\endgroup$ – Mike V.D.C. Mar 27 at 4:02
  • $\begingroup$ I am asking about if precomposition with every continuous function from your base spaces are continuous, not just an inclusion map. $\endgroup$ – Liquid Mar 27 at 4:16
  • $\begingroup$ You should add your proof for the path connected case. $\endgroup$ – Paul Frost Mar 27 at 16:52
  • $\begingroup$ I added it but while I was writing it up I realized it only really works if the paths in $X\times Y$ can be sufficiently short. $\endgroup$ – Liquid Mar 27 at 18:17
  • $\begingroup$ @Liquid Your function constructed by "connecting paths between $z_n$ and $z_{n+1}$" need not be continuous at $0$. I assume that you chunk $[0,1]$ into $[1/(n+1), 1/n]$ intervals and take path at each one? Assume that the glueing is continuous. Let $\lambda_n:[1/(n+1), 1/n]\to X$ be such paths. Note that any sequence $x_n$ such that $x_n\in[1/(n+1), 1/n]$ converges to $0$. And thus the sequence $\lambda_n(x_n)$ has to converge to $z_\infty$ by continuity. With that property you can easily constuct a counterexample: paths with mid-points converging somewhere else. $\endgroup$ – freakish Mar 27 at 22:01


$$X=\Bbb R \text{ with the usual topology};$$ $$Y=\Bbb R \text{ with the cocountable topology };$$ $$Z=\Bbb R \text{ with the cofinite topology};$$ $$f:X\times Y\rightarrow Z$$ $$f(x,y)= \begin{cases} \frac{xy}{x^2+y^2}, & \text{if } (x,y)\neq(0,0) \\[2ex] 0, & \text{if } (x,y)=(0,0) \end{cases} $$

Then $f$ is not continuous, otherwise, $f$ restricted to the diagonal $\Delta$ of $X\times Y$ will be continuous. But $f(x,x)=\frac{1}{2}$ for $(x,x)\neq (0,0)$; Thus $(f|_{\Delta})^{-1}(\frac{1}{2})=\Delta \setminus \{(0,0)\}$, which is not closed in $\Delta$, though $\{\frac{1}{2}\}$ is closed in $Z$.

Lemma 1:

Every continuous function $g:\Bbb R \to \Bbb R_{\text{cocountable}}$ must be constant.

proof: $f(\Bbb Q)$ is countable, hence closed in $\Bbb R_{\text{cocountable}}$. $f^{-1}(f(\Bbb Q))$ is a closed set containing $\Bbb Q$, hence equal to $\Bbb R$. $f^{-1}(f(\Bbb Q))=\Bbb R$ implies $f(\Bbb R)=f(f^{-1}(f(\Bbb Q)))\subseteq f(\Bbb Q) $, hence $f(\Bbb R)$ is countable. A countable, connected subset of $\Bbb R_{\text{cocountable}}$ must be a one-point set. Hence $f$ is constant. $\square$

Claim 1:

For every continuous function $g: \Bbb R \to \Bbb R\times \Bbb R_{\text{cocountable}}$, the image must be contained in $\Bbb R\times \{y_0\}$ for some $y_0$.

proof: $\pi_2\circ g$ is a continuous function from $\Bbb R$ to $\Bbb R_{\text{cocountable}}$. ($\pi_2$ is the projection onto the second coordinate). By lemma 1, $\pi_2\circ g$ is constant. Hence the result follows. $\square$

Hence for every continuous function $g: \Bbb R \to \Bbb R\times \Bbb R_{\text{cocountable}}$, the precomposition $f\circ g:\Bbb R\to \Bbb R_{\text{cofinite}}$ is equal to $f(x,y_0)=\frac{x^2y_0^2}{x^2+y_0^2}$ ($y_0$ fixed). Replacing $\Bbb R_{\text{cofinite}}$ by the finer topology $\Bbb R$, the function is easily seen to be continuous by standard calculus argument. Hence $f\circ g:\Bbb R\to \Bbb R_{\text{cofinite}}$ is continuous.

Lemma 2:

Every continuous function $h:\Bbb R_{\text{cocountable}} \to \Bbb R$ must be constant.

proof: (continuity on co-countable topology drhab's answer) Let $h$ be such a function with e.g. $0,1\in h\left(\mathbb{R}\right)$. Suppose $f$ is continuous and let $D_0,D_1$ be disjoint open sets containing $0$ and $1$ respectively. Then $h^{-1}\left(D_0 \right)$ and $h^{-1}\left(D_1\right)$ must be disjoint sets both having a countable complement. Then $h^{-1}\left(D_0\right)$ as a subset of the complement of $h^{-1}\left(D_1\right)$ is countable and consequently $\mathbb{R}=h^{-1}\left(D_0 \right)\cup h^{-1}\left(D_0\right)^{c}$ is countable. Contradiction. $\square$

Claim 2:

For every continuous function $h: \Bbb R_{\text{cocountable}} \to \Bbb R\times \Bbb R_{\text{cocountable}}$, the image must be contained in $\{x_0\}\times\Bbb R_{\text{cocountable}}$ for some $x_0$.

proof: $\pi_1\circ h$ is a continuous function from $\Bbb R_{\text{cocuntable}}$ to $\Bbb R$. ($\pi_1$ is the projection onto the first coordinate). By lemma 2, $\pi_1\circ h$ is constant. Hence the result follows. $\square$

Hence for every continuous function $h: \Bbb R_{\text{cocountable}} \to \Bbb R\times \Bbb R_{\text{cocountable}}$, the precomposition $f\circ h:\Bbb R_{\text{cocountable}}\to \Bbb R_{\text{cofinite}}$ is equal to $f(x_0,y)=\frac{x_0^2y^2}{x_0^2+y^2}$ ($x_0$ fixed). $(f\circ h)^{-1}(A)$ is closed (i.e. countable) for every closed (i.e. finite) set $A$. In fact, the preimage of each element can have cardinality at most $2$ (by looking at the $\text{graph}^1$). Hence $f\circ h:\Bbb R_{\text{cocountable}}\to \Bbb R_{\text{cofinite}}$ is continuous.

$^1$enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.