Prove $f : A\rightarrow B, g: B\rightarrow C$ , and $g\circ f: A \overset{1-1}{\rightarrow}C$, then $f:A\overset{1-1}{\rightarrow}B$ Could anyone please explain how to approach this problem, I'm honestly having a hard time figuring out where to start the problem. I know that I have to show that $\forall x,y\in A$ , if $f(x)=f(y)$, then $x=y$. 
Therefore, my educated guess would be to start by letting $x,y\in A$ s.t. $f(x)=f(y)$
 A: Look at what you’re trying to prove: that $f:A\to B$ is one-to-one. Now use the definition of one-to-one to reduce this to a more elementary statement to prove: you want to show that if $x,y\in A$, and $f(x)=f(y)$, then $x=y$. That’s what it means for $f$ to be one-to-one. The most straightforward approach, therefore, starts like this:

Let $x,y\in A$ be arbitrary, and suppose that $f(x)=f(y)$; we want to show that $x=y$.

Now what do we know? We know that $g\circ f$ is one-to-one. Is there any way to apply this? First we have to get something involving $g\circ f$. That’s easy: apply $g$ to both sides of $f(x)=f(y)$.

Clearly $g\big(f(x)\big)=g\big(f(y)\big)$, i.e., $(g\circ f)(x)=(g\circ f)(y)$.

Now use the fact that $g\circ f$ is one-to-one to conclude ... what?
A: Choose $x, y \in A$ subject to $f(x) = f(y)$. Take $g$ on both sides to get
$$g(f(x)) = g(f(y))$$
Writing this as composition gives
$$(g\circ f)(x) = (g \circ f)(y)$$
What can you now conclude, knowing that $g\circ f$ is 1-1?
A: We need to prove for $x,y\in A$:
(1) $f(x)=f(y)\implies x=y$,
as you correctly observe. However,
(2) $f(x)=f(y)\implies g(f(x))=g(f(y))$ (this is always true for maps $f:A\to B$ and $g:B\to C$).
Can you now use the injectivity of $g\circ f:A\to C$ to conclude (1) from (2)?
In general, when you're given information such as that above about an "external map" $g:B\to C$, then you should use that information. In this case, we needed to establish (1) and we did this by observing (2) which subsequently allowed use to use something we already know (the injectivity of $g\circ f:A\to C$) to reach our conclusion (1).
I hope this helps!
