Let $A$ be an abelian group and let $\mathbb{T} \subset \mathbb{C}$ be the unit circle. For $a,b \in A$, let $M(a,b) \subset \mathbb{T}^A$ be the set of all functions $f: A \to \mathbb{T}$ such that $f(a+b) = f(a) + f(b)$. Why is $M(a,b)$ closed?
Here $\mathbb{T}^A$ has the product topology.
I hope I do not misunderstand your question and by the addition $+$ on $\mathbb{T}$ you mean the usual multiplication via the identification of $\mathbb{T}$ with $\mathbb{R}/\mathbb{Z}$. If this is the case then note that because evaluation $\pi_a : \mathbb{T}^A \rightarrow \mathbb{T}, f \mapsto f(a)$ is continuous for the product topology, so is $$m_{a,b} : \mathbb{T}^A \rightarrow \mathbb{T}, f \mapsto f(a+b)- f(a) - f(b) = \pi_{a + b}(f) - \pi_a(f) - \pi_b(f)$$ as group multiplication and inversion on $\mathbb{T}$ are continuous. Now $M(a,b)$ is the inverse image of the identity element in $\mathbb{T}$ under $m_{a,b}$. As points are closed in $\mathbb{T}$, so is $M(a,b)$.