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In this link below, the asker provided answers to his question for critique. I'm confused about the answer to the question "(d)". Proof of properties of injective and surjective functions.

What I already know:
Now, since we want to prove that $g$ is injective, I assume that we want to prove if $g(a)=g(b) $, then $a =b$. I also know that we can prove the contrapositive. That is if $a\neq b$ then $g(a) \neq g(b)$. The asker starts by assuming $a\neq b$. So I guess they are going for the contrapositive. But what I don't get is the sentence after. he wrote:

if $(f\circ g)(a)\neq(f\circ g)(b)$, then $f(g(a))\neq f(g(b))$. Hence $g(a) \neq g(b)$.

From that 2 sentences, I think it is the same as saying $(f\circ g)(a)\neq(f\circ g)(b) \implies g(a) \neq g(b)$

but from the hypothesis , $f\circ g$ is injective. Which should mean $(f\circ g)(a) = (f\circ g)(b) \implies g(a) = g(b)$. But the asker use $(f\circ g)(a)\neq(f\circ g)(b)$ to conclude $g(a) \neq g(b)$. Isn't that like using $\neg p\implies \neg q$ to prove $p\implies q$. This is denying the hypothesis.

How did he conclude $g(a) \neq g(b)$. and what does $a\neq b$ have anything to do with "if $(f\circ g)(a)\neq(f\circ g)(b)$, then $f(g(a))\neq f(g(b))$" to begin with.

I am very inexperience, and I don't believe that the asker was wrong, so that is why I wrote out my thought process. Can someone point out what went wrong in my thought process?

Chhangsreng P
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  • If I have a function $h:A\to B$ and two elements $x,y\in A$ and I know that $h(x)\neq h(y)$ then surely $x\neq y.$ If $x$ and $y$ were the same element in $A$, then they have to map to the same element of $B!$ – D. Brogan Jan 23 '23 at 13:32
  • @D.Brogan Wait, WHAT. I never realize that the converse of the definition of injective function is also valid.

    OKay, now I understand that $(f\circ g)(a)\neq(f\circ g)(b) \implies g(a) \neq g(b)$ is valid. But how does that have anything to do with proving that g is an injective function? Why does $ a /neq b $ has anything to do with that statement in the first place?

     – Chhangsreng P Jan 23 '23 at 14:11

1 Answers1

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Just to be clear, in my comment the "converse of the definition of injective function" is just the statement that the function is well defined: every input only gives one output.

That being said, the proof really goes like this. Suppose $a\neq b.$ Then because $f\circ g$ is an injective function, we have $(f\circ g)(a)\neq (f\circ g)(b).$ But by definiton, $(f\circ g)(a) = f(g(a))$ and $(f\circ g)(a) = f(g(b)).$ So we actually know that $f(g(a))\neq f(g(b)).$ But $f$ is a function, so like any other function, if the outputs aren't the same, the inputs cannot be the same either; this is the content of my comment. That means that the inputs of $f$ in that (in)equation, in this case being $g(a)$ and $g(b),$ are not equal, i.e. $g(a)\neq g(b).$

So we started from $a\neq b$ and found that $g(a)\neq g(b).$ This proves that $g$ is an injective function.

D. Brogan
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  • Thank you so so much. I think my confusion is coming from the part that I didn't know that $a \neq b$ could imply $(f\circ g)(a) \neq f\circ g)(b)$. I didn't know that the inputs a and b work the same way that they do on a compose function as a normal function, and I didn't know that an injective composed function works the same way as a normal function. – Chhangsreng P Jan 23 '23 at 18:47