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Let $G$ be a topological group. For each $n \in \mathbb{Z}$, consider the continuous functions $f_{n} \colon G \to G : x \mapsto x^{n}$, and set $F := \{f_{n} \mid n \in \mathbb{Z}\}$.

Is there a nice/easy description of the closure of the set $F$ in $C(G,G)$ with respect to the topology of pointwise convergence? And what is the induced subspace topology on this closure?

Example: If $G$ is the (topological) circle group, then $F = Hom(G,G)$ is closed in $C(G,G)$ with respect to the topology of pointwise convergence. However, I do not know how the induced subspace topology on $F$ can be described. (It is known from Pontrjagin Duality that compactness of $G$ implies that the compact-open topology on $F$ coincides with the discrete topology, but this does not tell anything about the topology of pointwise convergence.)

Ideally, I would like to have a general statement, e.g, for compact Hausdorff groups. Does anybody know a suitable reference?

Edit:

I originally stated the example differently, namely

Example: If $G$ is the group associated to one-dimensional sphere, then $F$ is closed in $C(G,G)$ and the topology of pointwise convergence on $F$ coincides with the discrete topology.

However, in this example, the topology of pointwise convergence on $F$ does probably not coincide with the discrete topology. The proof that I had mind requires this topology to be frist-countable on $F$, but I cannot find an argument for this. In general, topological spaces with countable underlying set need not be first-countable (see Arens-Fort Space), so I guess the second statement of the example is just not true.

Let $G$ be a topological group. For each $n \in \mathbb{Z}$, consider the continuous functions $f_{n} \colon G \to G : x \mapsto x^{n}$, and set $F := \{f_{n} \mid n \in \mathbb{Z}\}$.

Is there a nice/easy description of the closure of the set $F$ in $C(G,G)$ with respect to the topology of pointwise convergence? And what is the induced subspace topology on this closure?

Example: If $G$ is the (topological) circle group, then $F = Hom(G,G)$ is closed in $C(G,G)$ with respect to the topology of pointwise convergence. However, I do not know how the induced subspace topology on $F$ can be described. (It is known from Pontrjagin Duality that compactness of $G$ implies that the compact-open topology on $F$ coincides with the discrete topology, but this does not tell anything about the topology of pointwise convergence.)

Ideally, I would like to have a general statement, e.g, for compact Hausdorff groups. Does anybody know a suitable reference?

Edit:

I originally stated the example differently, namely

Example: If $G$ is the group associated to one-dimensional sphere, then $F$ is closed in $C(G,G)$ and the topology of pointwise convergence on $F$ coincides with the discrete topology.

However, in this example, the topology of pointwise convergence on $F$ does not coincide with the discrete topology. The proof that I had mind requires this topology to be frist-countable on $F$, but I cannot find an argument for this. In general, topological spaces with countable underlying set need not be first-countable (see Arens-Fort Space), so the second statement of the example is just not true.

Let $G$ be a topological group. For each $n \in \mathbb{Z}$, consider the continuous functions $f_{n} \colon G \to G : x \mapsto x^{n}$, and set $F := \{f_{n} \mid n \in \mathbb{Z}\}$.

Is there a nice/easy description of the closure of the set $F$ in $C(G,G)$ with respect to the topology of pointwise convergence? And what is the induced subspace topology on this closure?

Example: If $G$ is the (topological) circle group, then $F = Hom(G,G)$ is closed in $C(G,G)$. It is known from Pontrjagin Duality that compactness of $G$ implies that the compact-open topology on $F$ coincides with the discrete topology. However, I do not know how the topology of pointwise convergence on $F$ can be described.

Ideally, I would like to have a general statement, e.g, for compact Hausdorff groups. Does anybody know a suitable reference?

Edit:

I originally used another example above, namely

Example: If $G$ is the group associated to one-dimensional sphere, then $F$ is closed in $C(G,G)$ and the topology of pointwise convergence on $F$ coincides with the discrete topology.

However, in this example, the topology of pointwise convergence on $F$ does probably not coincide with the discrete topology. The proof that I had mind requires this topology to be frist-countable on $F$, but I cannot find an argument for this. In general, topological spaces with countable underlying set need not be first-countable (see Arens-Fort Space), so I guess the second statement of the example is just not true.

Let $G$ be a topological group. For each $n \in \mathbb{Z}$, consider the continuous functions $f_{n} \colon G \to G : x \mapsto x^{n}$, and set $F := \{f_{n} \mid n \in \mathbb{Z}\}$.

Is there a nice/easy description of the closure of the set $F$ in $C(G,G)$ with respect to the topology of pointwise convergence? And what is the induced subspace topology on this closure?

Example: If $G$ is the (topological) circle group, then $F = Hom(G,G)$ is closed in $C(G,G)$ with respect to the topology of pointwise convergence. However, I do not know how the induced subspace topology on $F$ can be described. (It is known from Pontrjagin Duality that compactness of $G$ implies that the compact-open topology on $F$ coincides with the discrete topology, but this does not tell anything about the topology of pointwise convergence.)

Ideally, I would like to have a general statement, e.g, for compact Hausdorff groups. Does anybody know a suitable reference?

Edit:

I originally stated the example differently, namely

Example: If $G$ is the group associated to one-dimensional sphere, then $F$ is closed in $C(G,G)$ and the topology of pointwise convergence on $F$ coincides with the discrete topology.

However, in this example, the topology of pointwise convergence on $F$ does probably not coincide with the discrete topology. The proof that I had mind requires this topology to be frist-countable on $F$, but I cannot find an argument for this. In general, topological spaces with countable underlying set need not be first-countable (see Arens-Fort Space), so I guess the second statement of the example is just not true.