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

[Ozma]

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

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

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

 

[blamocur]

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

 

[Ozma]

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

 

[blamocur]

@blamocur: Thank you for the help, I didn't mean that [imath]g(f(a_1))=g(f(a_2))[/imath] because of the injectivity but that the implication [imath]g(f(a_1))=g(f(a_2)) \implies a_1=a_2[/imath] comes from the injectivity. Have I wrote something wrong that made you think that [imath]g(f(a_1))=g(f(a_2))[/imath] comes from [imath]g\circ f[/imath] 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 "[imath]\Longrightarrow[/imath]".

 

[pka]

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

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]