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For nonempty sets $A$ and $B$ and functions $ f: A \to B$ and $g: B \to A$, suppose that $ g \circ f = i_A$. Prove that if $f$ is onto, then $g$ is $1-1$.

Here is what I started, I don't think it's correct.

Suppose $f$ is onto i.e for each element $b \in B$ there is an element of $a\in A$, $f(a)=b$. (definition of onto).

Let $y,\ z$ be any two elements of $B$.Then $y=f(a)$ and $z=f(b)$ for some $a,b \in A$.

If $g(y)=g(z)$, then $g(f(a)) = g(f(b))$,which implies that $a=b$, so $f(a)=f(b)$, so $y=z$.

Thus, whenever $g$ maps $y$ and $z$ to the same element in $A$, $y=z$, so $g$ is one-to-one.

PS: New to proof writing, any constructive criticism is much appreciated.

PinkyWay
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user19059
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  • I've done some editting. You're correct. I hope you know why $a=b$ – Hesky Cee Aug 09 '14 at 01:13
  • Thank you so much. a=b since g(f(a))=g(f(b)) is bijective. Please correct me if I'm wrong. – user19059 Aug 09 '14 at 02:02
  • @user19059 Actually, we have my assumption that $g \circ f$ is the identity map, so $g(f(x)) = x$ for all $x \in A$. So, since we found $g(f(a)) = g(f(b))$, but since $g \circ f$ is the identity map, we have $g(f(a)) = a$, and $g(f(b)) = b$, and since $g(f(a)) = g(f(b))$, this implies $a = b$. Hope that helps! – layman Aug 09 '14 at 02:19
  • I appreciate your help. @user46944 – user19059 Aug 09 '14 at 03:36

1 Answers1

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Your proof is correct, as someone else stated. I just wanted to post my version of the proof with all of the details laid out, because I personally think no details should be left out of a proof (otherwise, it is usually hard for me to follow it). Here is my version:

Let $f: A \rightarrow B$ and $g: B \rightarrow A$ be functions such that $g \circ f = i_{A}$. Suppose that $f$ is onto. Then we want to show that $g$ is one-to-one, that is, if $g(a) = g(b)$, then $a = b$.

Suppose $a, a' \in A$ such that $g(a) = g(a')$. We want to show $a = a'$. But since $a,a' \in A$ and $f$ is onto, then we know there exists $b, b' \in B$ such that $f(b) = a$ and $f(b') = a'$. Thus, $g(a) = g(a') \implies g(f(b)) = g(f(b'))$. But by assumption, $g \circ f$ is the identity map, so we have $g(f(b)) = b$, and $g(f(b')) = b'$. But this shows that $b = b'$ since $g(f(b)) = g(f(b'))$. Finally, since $f$ is by assumption a function, it is well-defined, so one input results in at most one output. Thus, since $b = b'$, they are the same input, which means $f(b) = f(b')$, and this implies $a = a'$, as desired.

layman
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