Here is a question, the content in red is the question and the underlined area was left blank to answer it. The diagram is made by me to help understand the question,
I am unable to get the point that how does it prove that it is one to one?
Here is a question, the content in red is the question and the underlined area was left blank to answer it. The diagram is made by me to help understand the question,
I am unable to get the point that how does it prove that it is one to one?
______ #1: "$g$ is one to one"
______ #2: "$f$ is one to one"
The meaning of "$g$ is one to one" is that $g(x_1)=g(x_2)\implies x_1=x_2$ for any $x_1,x_2\in B$.
In your case, we have assumed that $(g\circ f)(a_1)=(g\circ f)(a_2).$ In other words, $g(f(a_1))=g(f(a_2)).$ Thus, we have two elements $f(a_1),f(a_2)\in B$ such that $g(f(a_1))=g(f(a_2))$. By the assumption that $g$ is one to one, this implies that $f(a_1)=f(a_2)$.
Then use the same reasoning to conclude that $a_1=a_2$ because $f$ is one to one.
Thus, $(g\circ f)(a_1)=(g\circ f)(a_2)\implies a_1=a_2$ for any $a_1,a_2\in A$. Therefore the function $(g\circ f):A\to C$ is one to one.