# Is a function $X \times Y \to Z$ continuous iff precomposition with continuous functions $X\to X\times Y$ and $Y\to X\times Y$ are continuous

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 $$Z$$.

• 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! – Mike V.D.C. Mar 27 at 4:02
• I am asking about if precomposition with every continuous function from your base spaces are continuous, not just an inclusion map. – Liquid Mar 27 at 4:16
• You should add your proof for the path connected case. – Paul Frost Mar 27 at 16:52
• 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. – Liquid Mar 27 at 18:17
• @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. – freakish Mar 27 at 22:01

Let

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

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.

$$^1$$