Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have a map $f:X \times Y \rightarrow Z$, where $X,Y$, and $Z$ are topological spaces. Are there any conditions on $X$,$Y$, and $Z$ that would allow one to determine that $F$ is continuous if it was known that it was continuous in each variable? It seems like there should be a theorem related to this.

By definition, a path homotopy $F: X \times I \rightarrow Y$ is continuous. What results in algebraic topology would not hold if we only required the map to be continuous in each variable? Would path homotopies not necessarily generate the fundamental group?

share|cite|improve this question
Even in the nice case $X=Y = \mathbb R$ the continuity in each variable is not sufficient for the continuity as a function of two variables - the same will hold if you take $X=Y=[0,1]$. – Ilya Jul 13 '11 at 20:43
f(x,y)=$ \frac{2xy}{x^2+y^2}; f(0,0)=0 $is a standard counterexample; continuous for each of x,y separately , but not continuous (check the limit at $(0,0)$ , e.g., along the direction y=x ). You may have to use sequential continuity ( if you have first-countability), or just the "old-fashioned" way, by showing that the inverse image under f of an open set in Z is open in the product topology of $X \times Y \$ – gary Jul 13 '11 at 21:10

Letting $S^1=\{z\in\mathbb{C}\colon\vert z\vert=1\}$ be the unit circle, consider the map $F\colon S^1\times I\to S^1$ given by $$ F(e^{2\pi\theta i},s) = e^{2\pi\theta^si} $$ for $0 < \theta\le 1$ and $s\in I$. Then $F$ is continuous in each variable, $F(z,1)=z$ and $F(z,0)=1$. So, if you only required continuity in the individual variables, the circle would be contractible. More generally, every topological space would have trivial fundamental group. Suppose that $X$ is a topological space and $\gamma\colon I\to X$ is a closed curve. Define $F\colon I\times I\to X$ by $F(x,s)=\gamma(x^s)$ for $x,s\in I$ and $s > 0$, and $F(x,0)=\gamma(0)=\gamma(1)$. Then $\gamma$ is null-homotopic (relative to ${0,1}$). So, the fundamental group collapses to the trivial group, as do all the higher homotopy groups.

share|cite|improve this answer
There's a nice paper in the Math Monthly about this idea: Deloup, Florian "The fundamental group of the circle is trivial." Amer. Math. Monthly 112 (2005), no. 5, 417–425 – Dan Ramras Jul 14 '11 at 5:43

As is the case with two others who have responded (at the time I wrote this), I don't have an answer to your specific questions (conditions on $X$, $Y$, and $Z$; homotopy analogs for separately continuous maps). However, Piotrowski's 1996 survey paper or my 2005 sci.math post (which contains some references not given in Piotrowski's paper -- [2], [4], and [6]) might have something of interest to you or lead you to a relevant reference.

Zbigniew Piotrowski, "The genesis of separate versus joint continuity", Tatra Mountains Mathematical Publications 8 (1996), 113-126. [MR 98j:01026; Zbl 914.01007]

sci.math -- "Continuity in each variable vs. joint continuity" (4 June 2005)

share|cite|improve this answer

Your Answer


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