For any entourage $U,V$  there's an entourage $W$ such that $U\circ W\subseteq V\circ U$ Let $(X,\mathcal U)$ be a uniform space and let $U\in \mathcal U$. Is this statement true?
$$\forall V\in \mathcal U, \exists W\in \mathcal U, U\circ W\subseteq V\circ U$$
I think if the above statement is true then we can easily prove:
$$\forall V\in \mathcal U, \exists W\in \mathcal U, W\circ U\subseteq U\circ V$$
 A: The second statement follows from the first one by passing to inverses (i.e. reflecting along the diagonal).
Now consider the uniform structure on $\mathbb{R}$ consisting of all subsets of $\mathbb{R}^2$ that contain an open neighborhood of the diagonal. 
Just to avoid confusion with left and right, I want to use the notation
$U\circ V = \{(x,z)| \exists y: (x,y)\in U \wedge (y,z)\in V\}.$
Let $U:=\{(x,y)| |y-x|<1 \}\cup  \mathbb{R}\times \{0\}$.
$V:=\{(x,y)| |y-x|<1 \}$. 
Then $V\circ U=\{(x,y)| |y-x|<2 \}\cup \mathbb{R}\times \{0\}$. 
Now let $W$ be any entourage and let $Z$ be some open neighborhood of $0$ such that $\{0\}\times Z\subset  W$. Then $U\circ W$ will contain $\mathbb{R}\times Z$. Thus $U\circ W$ will not be contained in $V\circ U$. Since $W$ was arbitrary this should be a counterexample.
A: I recently found an answer to a similar question. Suppose:
$$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$
By axiom of choice,for each $V \in \mathcal U$, there's some symmetric $D_V\in \mathcal U$ such that
$$D_V\circ D_V\subseteq  V$$
and there's some symmetric $W_V\in \mathcal U$ such that:
$$U\circ W_V \subseteq D_V\circ U$$
and
$$W_V\subseteq D_V$$
so
$$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$
Therefore
$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$
Now see this thread.
