Prove that set of all functions from A to B forms topological group

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Hello, I was trying to prove part (b) of problem given above. But I am stuck there. My idea was to use definition of product topology. For example if $U$ is open set in $B^A$ then its projections are open too. On the other hand inverse image in respect to product map is set of all $(f,g)$ such that $fg \in U$. My idea is trying to prove that if I take the set of all such $f$s, lets say $F$, and all such $g$s, lets say $G$, that all projections of $F$ and $G$ should be open. Unfortunatly I can find connection between $U$ and those projections. Can you tell me does this make sense or give some other idea on how to do this?