My question seems far too basic to be unknown, but I could not find anything relevant...

Let $X$, $Y$ and $Z$ be compact connected metric spaces, and let $F \subset X \times Y$ and $G \subset Y \times Z$ be closed binary relations such that both projections are surjective. Suppose that they are also connected as topological spaces.

Is it true that $G \circ F$ is connected?

Probably I should also mention that a combinatorial version of this statement for connected graphs is known to be true.


No. There are two (discontinuous) surjective maps $f,g:S^1\to S^1$ whose graphs are connected but the graph of $g\circ f$ (as well as its closure) is not.

The map $f$ is defined as follows, using the standard parametrization of $S^1$ by $\mathbb R/2\pi\mathbb Z$. It is the identity on the complement from the arc (parametrized by) $[\pi/2,\pi]$. The arc $[\pi/2,\pi]$ is slightly stretched from the point $\pi/2$ so that its image is the arc $[\pi/2,\pi+\varepsilon]$. The map $g$ is similar but with a discontinuity at 0 rather than $\pi$.

The graphs of $f$ and $g$ can be parametrized by an interval and hence connected. The graph of $g\circ f$ is not, because the map has two simple discontinuities (at 0 and $\pi$) that divide the circle into two components.

The graphs are not closed in $S^1\times S^1$, but adding points $(\pi,\pi)$ and $(0,0)$ fixes this issue.

  • $\begingroup$ Oh, thank you. So I was very much wrong about graphs too. I thought it would follow easily from the amalgamation property for linear orders, but apparently it does not... $\endgroup$ – Alexander Shamov Nov 20 '12 at 11:29

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