# Continuity of Multiplication by Fixed Element

This is likely a simple question that I'm just missing, but nothing immediately came to mind.

When dealing with topological monoids, it is necessary to prove that the group operation is continuous. I.e. if we have a monoid $(G,\ast,e)$ with some topology $\tau$, then it is necessary that $-\ast-: G\times G\rightarrow G$ be continuous with respect to $\tau$ in order for $G$ to be a topological group.

Slightly related to this is in the consideration of the function $f_g:G\rightarrow G$ defined by $f_g(g')=g\ast g'$. Now, if the monoid operation is continuous, each $f_g$ will clearly be continuous by pre-composition with the product of the constant function sending all elements of $G$ to $g$ (which is always continuous) and the identity on $G$ (again, always continuous), and using the fact that the product of continuous maps is continuous. In the case where $G$ is a topological group, then $f_g$ is furthermore a homeomorphism.

However, it doesn't seem to me that the converse would necessarily hold true: if for every element $g$ of $G$, $f_g:G\rightarrow G$ defined as above is continuous, then $\ast$ is continuous.

This gets me to my question:

Is the converse true? If not, what are some counterexamples?

Consider for instance the one-dimensional vector space $\mathbb{R}$ with the co-compact topology (a non-empty set is open iff its complement is compact in the usual topology). In this topology, the space is $T_1$ (though not Hausdorff), [...] and the addition map $+: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is continuous in each coordinate (i.e. translations are continuous), but not jointly continuous; for instance, the set $\{ (x,y) \in {\mathbb R}: x+y \not \in [0,1]\}$ does not contain a non-trivial Cartesian product of two sets that are open in the co-compact topology.