I was wondering why the following claim is correct:
Let G* be the group of all continuous homomorphisms from the topological group G and the unit circle (call it T). Then G* is an intersection of a closed set in the point open topology of all functions from G to T and C(G,T) (set of all continuous functions from G to T).
G* is also called a character. Thanks in advance for all the help
The fact that a (continuous) function $f$ from $G$ to $T$ is a homomorphism can be expressed as a "closed condition", as it were, in the pointwise (product) topology on $T^G$. Namely, write for a fixed pair of $g,h \in G$:
$$A(g,h) = \{f \in T^G \mid f(g)\cdot f(h) = f(gh)\} = ((m_T \circ (\pi_g \Delta \pi_h)) \Delta \pi_{gh})^{-1}[\Delta_T]\text{,}$$
where $m_T$ is the multiplication map from $T \times T$ to $T$ (which is continuous) and for any group element $t \in G$, $\pi_t: T^G \rightarrow T$ is the projection onto the element $t$ (sending $f \in T^G$ to $f(t)$), which are all continuous by definition of the pointwise topology. Also, $\Delta_T = \{(x,x): x \in T\}$ is the diagonal in $T$ which is closed (as $T$ is Hausdorff), and $\Delta$ denotes the diagonal product of maps from a common domain (which produces continuous functions from continuous functions).
To expand a bit on this: $\pi_{gh}: T^G \rightarrow T$ is a projection, as are $\pi_g : T^G \rightarrow T$ and $\pi_h: T^G \rightarrow T$. The latter two are combined into a diagonal map $\pi_g \Delta \pi_h: T^G \rightarrow T \times T$, defined by $(\pi_g \Delta \pi_h)(f) = (f(g), f(h))$ (which is continuous as $\pi_g, \pi_h$ both are; a standard fact for product spaces). The latter is composed with $m_T$ so $(m_T \circ (\pi_g \Delta \pi_h))(f) = f(g)\cdot f(h) \in T$, and this is continuous as composition of continuous maps. Now, this last map is again combined as a diagonal product with the first-mentioned projection $\pi_{gh}$, so this total map $(m_T \circ (\pi_g \Delta \pi_h))\Delta \pi_{gh}$ sends $f$ to the pair $(f(g)\cdot f(h),f(gh))$. Which lies in $\Delta_T \subset T \times T$ iff the left hand side term equals the right hand side term.
So all sets $A(g,h)$ are closed and the homomorphisms in $T^G$ is just the set $\cap_{g,h \in G} A(g,h)$, which is also closed (in the pointwise topology) as the intersection of closed sets.