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  1. If $f \circ g$ is surjective, $f$ is surjective.
  2. If $f \circ g$ is surjective, $g$ is surjective.

$\textbf{Part 1:}$ Let $f:B \to A$ and $g:C \to B$. Assume $f \circ g$ is surjective. Since $f(g(x))$ is surjective, for all $a \in A$ there is a $c \in C$ such that $f(g(c))=a$. But since $g(c) \in C$ (by definition of g), that means for all $a \in A$, there is a $b \in B$ (namely g(c) such that f(b)=a). So f is surjective.

$\textbf{Part 2:}$ How would I amend the proof for Part 1 for Part 2?

Seth
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Username Unknown
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1 Answers1

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The point $2.$ is false.

Example : Take $f(x)=\min(1,2x),g(x)=x/2$ from $[0,1]$ to $[0,1]$. $(f\circ g)(x)=x$ is surjective, but $g$ is not.

This way is true, however: I call that the SI (S on the left, I on the right) lemma as a reminder :

If $f\circ g$ is S urjective, then $f$ is surjective ($f$ is on the left, S is on the left)

If $f\circ g$ is I injective, then $g$ is injective ($g$ is on the right, I is on the right)

Hippalectryon
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  • @Bryan oops thanks – Hippalectryon Sep 11 '14 at 22:32
  • I like the name 'SI lemma' though. I'm going to adapt it to 'EM lemma' for epimorphisms and monomorphisms. Thanks for a mnemonic. –  Sep 11 '14 at 22:34
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    An easier counterexample is $A={0}$, $B={1,2}$, $C={3}$; $f\colon B\to C$ where $f(1)=f(2)=3$, $g\colon A\to B$ where $g(0)=1$. Then $f\circ g$ is injective and surjective, but $g$ is not surjective and $f$ is not injective. – egreg Sep 11 '14 at 22:38
  • I was very confused about your SI lemma prior to your edit. It makes much more sense now :-) I'll see if my long-term memory is capable of accepting any new mnemonics... –  Sep 11 '14 at 23:41