Transitivity of 1-to-1 functions If $g:A\rightarrow B$ and $f:B\rightarrow C$ are both 1 to 1, how do we prove that $f\circ g$ is 1-to-1?
What type of a proof method would be required exactly?
 A: To show $h : A \to C$ is one-to-one you need to show that if $h(a)=h(a')$ then $a=a'$.
Do this with $h = f \circ g$, using the fact that both $g$ and $f$ are one-to-one and that $(f \circ g)(x) = f(g(x))$ for all $x$.
A: You want to show that if $(f\circ g)(a)=(f\circ g)(a')$ then $a=a'$.
So suppose that $(f\circ g)(a)=(f\circ g)(a')$.
Note that $(f\circ g)(a)=f(g(a))$. Similarly, $(f\circ g)(a')=f(g(a'))$.
If $f(g(a))=f(g(a))$, then, because $f$ is one to one, we have $g(a)=g(a')$. Now we are close to the end. 
The proof could also be written up as a proof by contradiction. Suppose that $(f\circ g)(a)=(f\circ g)(a')$.  We show that $a=a'$. Suppose to the contrary that $a\ne a'$. Then $g(a)\ne g(a')$, since $g$ is one to one. But then $\dots$.
Which version we use is a matter of taste. The two versions are essentially the same. 
A: The proof method depends on your definition of what 1-to-1 function is, suppose for $a,b\in A$ that $a\neq b$, then the injectivity of $g$ gives you that $g(a)\neq g(b)$ and the injectivity of $f$ gives you that $f(g(a))\neq f(g(b))$, but this implies that $f\circ g$ is injective.
A: If $f:A\to B$ is one-one so we know that: $$f(a)=f(b)\Longrightarrow a=b$$ First af all we see that $h=f\circ g:A\to C$ is a function as you noted and as you know this fact. Let $$h(a)=h(b)$$ then $$f\circ g(a)=f\circ g(b)$$ but $f$ is a one-one function so $$f\circ g(a)=f\circ g(b)\Longrightarrow g(a)=g(b)$$ but $g$ has the same property so, $$g(a)=g(b)\Longrightarrow a=b$$ Overall we reach to this: $$f\circ g(a)=f\circ g(b)\Longrightarrow g(a)=g(b)\Longrightarrow a=b$$ This means that $h$ is one-one function.
