# Sets, Functions and Exclusivity [closed]

If $$f: A \longrightarrow B$$ and $$g: B \longrightarrow A$$ are functions, $$A$$ and $$B$$ are sets and $$a_1, a_2, ..., a_n$$ in $$A$$, can the following ever be true for some $$a_n$$?

$$g(f(a_1)) \neq a_1$$

$$g(f(a_n)) = a_n$$

## closed as off-topic by Cesareo, Gibbs, José Carlos Santos, Aweygan, SBareSFeb 9 at 22:21

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## 1 Answer

I hope I understood the question since it feels like some info is missing. But yes if those conditions are the only ones, you can define $$f:A\to B$$ almost however you like and then $$g:B\to A$$ as $$g(b)=a_n$$ for all $$b \in B$$

• But if $b = a_1$, then $g(b) = a_1$ violates the claim. – Jossie Calderon Feb 9 at 19:33
• I think you mean if $b=f(a_1)$, if thats the case $g(f(a_1))=g(b)=a_n$ since for any $b$, $g$ is going to return you $a_n$ – Javi maxwell Feb 9 at 19:57
• Yes, I meant $b = f(a_1)$. But when $n = 1$, then $g(f(a1)) = g(b) = a_1$. That violates the claim. – Jossie Calderon Feb 9 at 20:23
• No, I am taking $a_n$ as a fixed value in $A$, $g$ is a constant function. In your question, you said that $A$ had a finite number of elements from $a_1$ to $a_n$. I'm using that last element $a_n$....if you meant to ask if $g(f(a_i))=a_i$ for $i=1,...,n$ while $g(f(a_1))≠a_1$ also holds, then no, the conditions contradict each other – Javi maxwell Feb 9 at 20:38
• if $i=1$ then $g(f(a_1))=a_1$ but $g(f(a_1))\neq a_1$, thats why they contradict each other – Javi maxwell Feb 11 at 5:54