Uniformly continuous maps between topological groups Let $G$ be a topological group. For every neighbourhood $U$ of the identity, let $L_U$ be the set of all pairs $(x,y) \in G \times G$ such that $x^{-1} y \in U$.
For topological groups $G$ and $H$, a map $f \colon G \to H$ is called left uniformly continuous if for every open neighbourhood $V$ of the identity in $H$ there exists an open neighbourhood $U$ of the identity in $G$ such that $(f(x),f(y)) \in L_V$ for all $(x,y) \in L_U$.
Is the map $G \times G \to G$, $(x,y) \mapsto x^{-1}y$ left uniformly continuous?
That is for any open neighbourhood $W$ of the identity does there exists neighbourhoods $U,V$ of the identity such that $(x_1^{-1}x_2, y_1^{-1}y_2) \in L_W$ whenever $(x_1,x_2,y_1,y_2) \in L_{U \times V}$.
If not, is there a nice example to show this is not true?
 A: The map is not generally left uniformly continuous. I'll begin with some remarks, then demonstrate the existence of a group where the map is not left uniformly continuous.
Note that $((x_1, y_1), (x_2, y_2)) \in L_{V \times W}$ if and only if $x_1^{-1} x_2 \in V$ and $y_1^{-1} y_2\in W.$ We need to use these conditions to force $(x_1^{-1}y_1, x_2^{-1}y_2) \in L_U,$ which is equivalent to $\phi((x_1, y_1), (x_2, y_2)) = y_1^{-1}x_1 x_2^{-1} y_2 \in U.$ Now, a general element of $L_V$ is of the form $(x, xv)$ where $v\in V$ and $x$ is any element of $G$, and similarly $(y, yw)$ is a general element of $L_W$. Furthermore $\phi((x, y), (xv, yw)) = y^{-1} x v^{-1} x^{-1} y w$ so $$\phi(L_{V\times W}) = \bigcup_{x\in G} x V^{-1} x^{-1} W \text{ and } \bigcup_{x\in G} x V^{-1} x^{-1} \subseteq \phi(L_{V\times W}),$$ and therefore we need to find some open neighborhood of the identity, $V$, such that all its conjugates are contained in an arbitrary open set. In general, this is not possible, which we demonstrate by looking at a particular (non compact) group.
Let $G$ be the group of isometries of $\mathbb{R}^2$ with the Euclidean metric, so $G$ is the group generated by rigid rotations and translations of the plane (and lets not have reflections, though this makes no real difference, its a separate connected component of the group). I claim that for all open neighborhoods of the identity $V$, and all positive real numbers, $l$,  there is some translation $t\in G$ and $v\in V$ such that $d(t v t^{-1} (\vec{0}), \vec{0}) = l$. To see this, note that $V$ contains some rotation, $v$, that fixes $\vec{0}$ and rotates the plane by $\epsilon >0.$ Then $t(a, b) = \left( a + \frac{l}{2\sin(\epsilon/2)}, b \right)$ is as required. In fact, for every point in the plane, there is an element of $\{ tvt^{-1}: t \text{ is a translation}\}$ that maps the origin to that point.
Therefore, if we let $U$ be the set of isometries that move the origin by strictly less than $1$, no matter what neighborhoods, $V$ and $W$, of the identity we choose, $\phi(L_{V\times W}) \not\subseteq U.$
I think that's all correct, now. This pdf was quite helpful to get me on the right track, especially lemma 4.2.
