Suppose you have functions $f$ and $g$ such that $f: Y \rightarrow T$ and $g: X \rightarrow Y$. I having a tough time understanding this?  I've learned about onto and 1 to 1 functions from a basic perspective, but I'm having a hart time understanding where to even start with this.  
(a) Suppose (f o g) is onto.  Claim g is onto. Prove or Disprove.
(b) Suppose (f o g) is onto.  Claim f is onto. Prove or Disprove.
(c) Suppose (f o g) is one to one.  Claim g is one to one.  Prove or Disprove 
(c) Suppose (f o g) is one to one.  Claim f is one to one.  Prove or Disprove 
We've been given a hint of one is true for onto, and one is true for one to one. 
 A: The function $f\circ g$ has domain $X$ and codomain $T$, i.e., $f\circ g: X \rightarrow T$.
For question (a), we are given that $f\circ g$ is onto. This means that everything in $T$ gets mapped to, i.e. for all $t \in T$, there exists some $x\in X$ such that $f\circ g(x) =t$. 
Can this happen without $g$ being onto? Pause here and try out some example functions, they can be functions on the reals or functions between small sets. Pause here before reading on.
Unpause: Were you able to find a counterexample? If not, here's a hint: let $T$ have only one element. Remember: a counterexample is all you need to disprove something.
For (b), we again have that $f\circ g$ is onto. However, we are now asked if $f$ must be onto? In this situation, start by writing an example where $f$ is not onto, and see if you can make $f\circ g$ onto. Once you've convinced yourself this is a futile task, you're ready to move on to proving that $f$ must be onto. 
We can do a proof by contradiction: we show that if $f$ is not onto, then $f\circ g$ is not onto [you can think of this as the contrapositive too, I'm not sure which you're more comfortable with, but it's effectively the same here]. If $f$ is not onto, that must mean there is an element $t' \in T$ to which no element in $Y$ maps under the function $f$.  I'll leave you here: Given what we know, could we possibly find an element in $X$ that gets mapped to some element in $Y$ by $g$ and then gets mapped to $t'$ by $f$? 
A: If (f o g) is onto, that means that for any x value that enters into g, there is a resulting value of g AND that each of these g values has a corresponding f output. This means that f's outputs are the ones that must have be fulfilled by all of the given inputs. This means that between (a) and (b), the claim that f is onto can be proven, where g does not have to be onto. 
However, if (f o g) is one to one, that means that each x value will map to exactly one value of g, and these g values will in turn each map to exactly one value of f. If any inputs were to map to the same g value, then these inputs would also map to the same f value and the composition would not be one to one, thus g must be one to one and (c) must be true. However, f could have inputs that are not members of g and are not one to one; this would not prevent the composition (f o g) from being one to one - so (d) cannot be proven. 
This does not provide proofs, but I hope that it does help you make sense of how one to one and onto functions apply with composed functions. 
