Why metrizable group requires continuity of inverse? A metrizable group is a metric space $(G,d)$ with a binary operation $\cdot$ such $(G,(\cdot))$ is a group and maps $(\cdot):G\times G\to G$ and $f:G\to G$ given by $(\cdot)(x,y)=xy$ and $f(x)=x^{-1}$ are continuous respect $d$.
Why requires definition that $f$ be continuous? Is it possible to have continuity of $(\cdot)$ without continuity of $f$?
 A: Take $G=\mathbb R$ with addition and put on it the topology which has the set $\{(-\infty,a):a\in\mathbb R\}$ as a basis. Then addition is continuous but inversion is not.
You want it metrizable, though...
A: I made a mistake here before. Sorry!
Continuity of multiplication is not sufficient for general topological groups.
Consider the standard additive group of reals with Sorgenfrey topology.
Inverse is clearly not continuous, since $-[a,b)=(-b,-a]$.
Multiplication (or, rather, addition) is continuous, since, intuitively, the open sets in the product topology are the sets which do not have "upper" and "right" edge (but may possibly have "left" and "lower" edge, or some parts of those). More precisely, for any $x+y=c\in [a,b)$, then let $0<\varepsilon<(b-c)/2$. Then for any $(x',y')\in [x,x+\varepsilon)\times [y,y+\varepsilon)$, $a\leq x+y\leq x'+y' <x+y+\varepsilon<b$, so $(+)^{-1}[a,b)$ is open.
EDIT:
I found an example in Topological Groups and Related Structures by Archangel'skii & Tkachenko, example 3.5.6 on pages 175-176. It is the group of homeomorphisms of the space $\lbrace n,0,1/n\mid n\in \mathbf N\rbrace$ with the natural topology, the compact-open topology.
It is metrizable, but not metrizable by a left- or right- invariant metric, and the inverse is not continuous.
