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Let $f : A \to B$ and $g : B \to C$ be functions.

If $g \circ f$ is injective and $f$ is surjective, then $g$ is injective.

This question that correctly proves this theorem, but my proof seems to be different, so I'd like clarification on whether it's correct.

My Proof

We want to show that, for any $b_1, b_2 \in B$, $g(b_1) = g(b_2)$ iff $b_1 = b_2$.

Let $g(f(a_1)) = g(f(a_2))$, where $a_1, a_2 \in A$.

$\therefore f(a_1) = b_1$ and $f(a_2) = b_2$, where $b_1, b_2 \in B$. (Since $f$ is surjective.)

$\therefore g(b_1) = g(b_2)$, where $b_1 = b_2$. (Since $g \circ f$ is injective.)

$\therefore$ $g$ is injective. $\ \ \ Q.E.D.$

I would greatly appreciate it if people could please review my proof and elaborate on anything I've done incorrectly.

The Pointer
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  • The phrasing is a bit awkward, for instance when you say "$g(b_1)=g(b_2)$, where $b_1=b_2$". – 57Jimmy Apr 03 '18 at 17:13
  • @DietrichBurde Oh? It looked different to me. So my proof is also correct? – The Pointer Apr 03 '18 at 17:14
  • @57Jimmy Awkward phrasing, but is the proof correct? And how would you suggest I improve the phrasing? – The Pointer Apr 03 '18 at 17:15
  • Why don't you just compare line by line with the proof by Andre at the duplicate? – Dietrich Burde Apr 03 '18 at 17:17
  • @ThomasAndrews consider $f: {1}\to\mathbb{Z}$, $f(1)=1$, and $g:\mathbb{Z}\to\mathbb{Z}$, $g(k)=1$. – TomGrubb Apr 03 '18 at 17:17
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    Yeah, misread the order. @ThomasGrubb – Thomas Andrews Apr 03 '18 at 17:18
  • I do not know if it is correct, because I do not know exactly what you mean: what is a consequence of what? Where are things defined? My improvement would coincide with the proof you have linked. Also, notice that the triple point is not just a bullet point, but a logical symbol that means "therefore" (maybe you used it with this in mind, but again, I am not sure, so I just wanted to point it out) – 57Jimmy Apr 03 '18 at 17:19

3 Answers3

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This is wrong because you are not starting with the hypothesis you are given. You want to show that $g(x) = g(y) \implies x = y$, so you have to start with the assumption that $g(x) = g(y)$. So begin with:

Suppose $g \circ f$ is injective and $f$ is surjective. We will show that $g$ is injective. Suppose $b_1, b_2 \in B$ with $g(b_1) = g(b_2).$ Then...

Jair Taylor
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There are the following main problems:

  • Start by saying "Assume that $g(b_1)=g(b_2)$. We want to prove that then $b_1=b_2$".
  • $a_1$ and $a_2$ are not defined in your argument, but you want them to be pre-images of $b_1$ and $b_2$, and they exist because $f$ is surjective. You should say this explicitly and before using them. For instance: "Since $f$ is surjective, there exist $a_1,a_2 \in A$ such that $f(a_1)=b_1, f(a_2)=b_2$."
  • Then follows the observation that $g\circ f(a_1) = g(b_1) = g(b_2) = g\circ f (a_2)$. Since $g \circ f$ is injective, this implies that $a_1 = a_2$.
  • From the equality $a_1 = a_2$ it follows directly that $b_1=f(a_1)=f(a_2)=b_2$. Hence the claim is proved.
57Jimmy
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1

Let's try:

1) Want to show that $g$ is injective.

Let $b_1,b_2 \in B$, and let $g(b_1)=g(b_2)$.

2) Since $f$ is surjective there are $a_1,a_2 \in A$ with

$f(a_1)=b_1$, and $f(a_2)=b_2$.

3)Consider $g \circ f:$

$g(f(a_1))= g(b_1)$, $g(f(a_2)) = g(b_2)$.

We have

$g(b_1)=g(b_2)$ (see point 1) above),

$g(f(a_1))=g(b_1)= g(b_2) = g(f(a_2)).$

4) Since $g \circ f$ is injective $a_1=a_2$.

5) Then $f(a_1)= f(a_2)$ , hence

$b_1 = f(a_1) = f(a_2) = b_2.$

6) We have shown:

$g(b_1)=g(b_2)$ implies $b_1=b_2$,

$g$ is injective.

Peter Szilas
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