Is this proof of g(f(x)) injective implies f injective correct?

[Ozma]

Let $f:A \to B$ and $g:B\to C$ be two functions, show that if $g\circ f$ is injective then $f$ injective.

My try: suppose that the theorem is false, that is suppose it is true that $g\circ f$ injective and$f$ not injective. Since is true that $f$ is not injective, it is true that there exist $a_1,a_2 \in A$ such that $f(a_1)=f(a_2)$ and $a_1 \ne a_2$.

By hypothesis it is true that $g \circ f$ is injective, hence it is true that $g(f(a_1)) = g(f(a_2)) \implies a_1=a_2$; so it is true that I have both $a_1=a_2$ and $a_1 \ne a_2$, a contradiction. Is this correct?

 

[blamocur]

Looks mostly right, but the fact that $g(f(a_1))=g(f(a_2))$ does not require $g\circ f$ to be injective.

 

[Ozma]

@blamocur: Thank you for the help, I didn't mean that $g(f(a_1))=g(f(a_2))$ because of the injectivity but that the implication $g(f(a_1))=g(f(a_2)) \implies a_1=a_2$ comes from the injectivity. Have I wrote something wrong that made you think that $g(f(a_1))=g(f(a_2))$ comes from $g\circ f$ injective? Sorry for my mistake in that case.

 

[blamocur]

@blamocur: Thank you for the help, I didn't mean that $g(f(a_1))=g(f(a_2))$ because of the injectivity but that the implication $g(f(a_1))=g(f(a_2)) \implies a_1=a_2$ comes from the injectivity. Have I wrote something wrong that made you think that $g(f(a_1))=g(f(a_2))$ comes from $g\circ f$ injective? Sorry for my mistake in that case.

You are right, my mistake: while reading the second paragraph I've grouped together the phrase before "$\Longrightarrow$".

 

[pka]

Suppose that $f(p)=f(q)$ then by definition $g\circ f(p)=g\circ f(q)$.
But we are given that $g\circ f$ is an injective so that it must be the case that $p=q$.
QED.

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