If $f$ is injective and $g$ is not injective, $g\circ f$ is not injective?

Question

If $f$ is surjective and $g$ is not surjective, $g\circ f$ is not surjective?

@JoséCarlosSantos I think g o f is not injective and is not surjective. — sdnaksn, Jan 14 '18 at 17:40
@sdnaksn Why do you think that $g \circ f$ is not injective nor surjective? — the_candyman, Jan 14 '18 at 20:11
@the_candyman because I know for sure that if f and g are both injective then g o f is injective, and if f and g are both surjective then g o f is surjective. So I was supposing that for this reason in this case g o f is not injective nor surjective — sdnaksn, Jan 14 '18 at 20:14
@DietrichBurde my question is about the composite function NOT BEING injective or surjective tho — sdnaksn, Jan 14 '18 at 20:18
Any combination has been solved here, see for example this question. — Dietrich Burde, Jan 14 '18 at 20:22
If $f$ is injective, and $g$ is not, then $g\circ f$ could still be injective. So answer is no. Did you read the duplicates on the right hand side? — Dietrich Burde, Jan 14 '18 at 20:27

score 0 · Answer 1 · answered Jan 14 '18 at 17:45

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Hint for the 1st question: $g$ can be injective in a smaller subset. Hint for the 2nd question: apply the definition.

answered Jan 14 '18 at 17:45

score 0 · Answer 2 · answered Jan 14 '18 at 17:46

What you conjectured in the comments is correct:

Take $f\colon\{1\}\longrightarrow\mathbb N$ and $g\colon\mathbb{N}\longrightarrow\mathbb N$ defined by $f(1)=1$ and $g(n)=1$ respectively. Then $g\circ f=f$, which is injective.
If $g$ is not surjective, then $g\circ f$ cannot be surjective, because if $f$ is a function from $A$ to $B$ and $g$ is a function from $B$ to $C$, if there is a $c\in C$ which cannot be expressed as $g(b)$ for some $b\in B$, then $c$ cannot be expressed as $g\bigl(f(a)\bigr)$ for soma $a\in A$ also. Otherwise, you would just take $b=f(a)$.

2 Answers2