Let V, W and Z be vector spaces, and let $T:V \rightarrow W$ and $U: W \rightarrow Z$ be linear. 
Let V, W and Z be vector spaces, and let $T:V \rightarrow W$ and $U: W \rightarrow Z$ be linear.
a).Prove if UT is one-to-one, then T is one-to-one. Must U also be one-to-one?
b). Prove if UT is onto, then U is onto. Must T also be onto?

for a). If UT is one-to-one, let $x \in V$, then $U(T(x))=0$ and $T(x)=0$. 

Question for a): Am I supposed to prove that x is also equal to 0? I am not sure how to get there. And I am shaky on whether U must be one-to-one. It seems like U must.

for b). If UT is onto, for $x \in V$, $U(T(x))= Z$
Since $T(x) \in W$, and image of U is equal to codomain, U is onto.

Question for b): I am not sure how to argue whether U must be onto. 

 A: Both holds also for functions that are not linear, so I will prove this in the traditional way. 
Recall that, if $f:A\to B$ is a function, $f$ is one-to-one if and only if for every $x$ and $y$ in $A$, 
$$f(x) = f(y) \textrm{ implies that } x=y.$$
Also, recall that $f$ is onto if and only if for any $b\in B$ there exists $a\in A$ such that $b = f(a)$.
So, for $(a)$, since our goal is to prove that $T$ is one-to-one, suppose that $u$ and $v$ are two vectors in $V$ such that $T(u) = T(v)$ and, in some way, we need to prove that $u=v$. But this is easy, just apply $U$ to both sides of the equation $T(u) = T(v)$ and use the fact that $UT$ is one-to-one.
For $(b)$, pick any $z\in Z$ and we need to find some vector $w\in W$ such that $z = U(w)$, right? Since $UT$ is onto, for $z\in Z$ there are $v\in V$ such that $z = (UT)(v) = U(T(v))$. What is supposed to be the vector $w$?
A: a. These are actually properties of regular set-maps, no linear algebra required. However, we can use the additional structure of linear maps to answer the question. Specifically, we can use the fact that $T$ is 1-1 if and only if $\text{ker}(T) = {0}. (I think this is what you're trying to do)
In order to do this, assume $v$ is non-zero and in the kernel of $T$. Where does $UT$ send $v$. Is this possible (remember that $UT$ is 1-1, and the above paragraph)
As for whether $U$ must be 1-1, try playing around with set-maps, and see if you can figure something out. 
b. Well what does it mean for a function $f$ to be onto? It means ``For all $y$ in the codomain, there is some $x$ so that $f(x) = y$." 
So, in order to prove something is onto, you start by saying ``let $y$ be in the codomain..." then go on to find an $x$ in the domain that maps to $y$. 
In this case, you have to use the fact that $UT$ is onto in order to produce the element $x$. 

I highly recommend working through some toy examples yourself. Again, these are properties of functions in general, not just linear maps. 
