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Given $X,Y,Z\neq \emptyset$
Let $f:X \rightarrow Y$ and let $g:Y \rightarrow Z$
I know that since $g\circ f$ is one-one therefore $g\circ f(x_1)=g\circ f(x_2)$ when $x_1=x_2$
In other words, $g(f(x_1))=g(f(x_2))$ when $x_1=x_2$
I don't know how to proceed further.

rschwieb
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Harsh Singh
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    Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. – José Carlos Santos Feb 13 '22 at 09:26
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    To be explicit, one should always put the question in the post body. It's OK for it to be in the title too, but not only there. – rschwieb Feb 16 '22 at 14:32

2 Answers2

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Assume that f is not one-one then there exist $x_1,x_2$ such that

$f(x_1)=f(x_2)$ and $x_1 \neq x_2$

$f(x_1)=f(x_2) \Rightarrow g(f(x_1))=g(f(x_2))$

However, since $gof$ is one-one

$x_1 \neq x_2 \Rightarrow g(f(x_1))\neq g(f(x_2))$

contradiction !! hence f must be one-one

Nevzat Eren Akkaya
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Hint: ''I know that since gof is one-one therefore gof(x1)=gof(x2) when x1=x2.''

No, it's vice versa: $g\circ f(x_1) = g\circ f(x_2)$ implies $x_1=x_2$.

Let $f(x_1) = f(x_2)$. Since $g$ is a function, $g\circ f(x_1) = g\circ f(x_2)$. Since $g\circ f$ is one-to-one, $x_1=x_2$. Hence, $f$ is one-to-one.

Wuestenfux
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