I'm trying to understand what exactly is going on in this example proof, but I am having trouble with the syntax of it. Here is the question, and the proof
Let $f : A\to B$ and $g : B \to C$ be functions.
Prove or disprove that if $g \circ f$ is onto, then $f$ is onto.
This statement is false, and here is a proof:
Let $A = C = \{ 1 \},$ $B = \{ 1, 2 \},$ $f = \{ (1, 2) \}$, and $g = \{(1, 1), (2,1)\}.$ Then, $g \circ f = \{(1,1)\},$ and is onto, but $f$ is not onto.
I have three main questions
1)Does $f$ mean that when you take the value $1$ from $A$, it maps to $2$ in $B$, and similarly with $g$, where $1$ and $2$ from $B$ maps to $1$ in $C$?
2) how does $g \circ f$ end up being $\{(1 , 1)\}$?
3) Why is $g \circ f$ onto, but $f$ not onto?
EDIT: By onto I mean, $\forall b \in B, $ $\exists a \in A$ such that $f(a) = b$
1)Does f mean that when you take the value 1 from A, it maps to 2 in B, and similarly with g, where 1 and 2 from B maps to 1 in C?
Yes, f and g act as you mentioned.
2) how does g∘f end up being {(1,1)}?
$gof$ is a function from $A$ to $C$
$$ gof(1)=g(f(1))=g(2)=1$$
3)Why is g∘f onto, but f not onto?
There is only one element $1$ in $C$ and it is covered by $gof$ thus $gof$ is onto.
$f$ is not onto because it does not cover $1$ which is in its codomain $B$