How do I read this question? (subject: bijections) Introduction
In Basic Algebra I, I am struggling with fully understanding the following exercise:

Show that $S\overset{\alpha}{\to}T$ is injective if and only if there is a map $T\overset{\beta}{\to}S$ such that $\beta\alpha=1_S$, surjective if and only if there is a map $T\overset{\beta}{\to}S$ such that $\alpha\beta=1_T$. In both cases, investigate the assertion: if $\beta$ is unique then $\alpha$ is bijective.

My Problem
I am struggling only with the bold portion. (I have written proofs by contradiction for the other aspects of the question.) What confuses me specifically is this:


*

*What is this question really asking? Is it saying, "What happens when $\beta$ is unique when both $\beta\alpha=1_S$ and $\alpha\beta=1_T$?" or is it saying, "What happens when $\beta$ is unique and either $\beta\alpha=1_S$ or $\alpha\beta=1_T$ is true?"


Remarks
As you can see, my real problem here is understanding precisely what is being asked. If it is asking, the first (both $\alpha\beta=1_T$ and $\beta\alpha=1_S$ are true), then we're simply constructing the very definition of a bijection. If it's asking the latter, I don't know what's going on . . . Are we somehow still constructing a bijection?
Can you all give me help on reading questions such as this?
 A: I parse this as follows:
The bold sentence refers separately to each of the two statements.
So expanded out, this would be:
1a.  Show that ... is injective if and only if ...  Show also that if $\beta$ is unique then $\alpha$ is bijective.
1b.  Show that ... is surjective if and only if...  Show also that if $\beta$ is unique then $\alpha$ is bijective.
A: Central Matter
With the help of Assad Ebrahim, I was able to figure out what's being said here. The crux is this:
$$
\beta \text{ is unique} \iff |S|=|T| \iff \alpha \text{ is bijective},
$$
which quickly implies that if $\alpha$ is bijective in either case.
Elaboration and Specifics
For the mapping $\alpha$ wherein $\beta\alpha=1_S$, we have that $\alpha$ is injective. However, $\beta$ can be any map $T\to S$ such that all the elements of $T$ map to $S$. This means that $\beta$ acts not just on the elements $\alpha(s)$; it acts on all elements of $T$. As a result, there are, in general, elements in $T$ which can be mapped to any $s\in S$. Thus, there are many possible maps $\beta$; we can create a new one simply by changing what a given $t\in T$ which is not $\alpha(s)$ maps to.
Now, the issue is this: We are supposing $\beta$ is unique. This means there cannot be elements in $T$ which are not equal to $\alpha(s)$. If there were, then $\beta$ would cease to be unique for the reason outlined just above. Hence, we have that $\alpha$ is also surjective. Therefore, $\alpha$ is bijective. $\blacksquare$
For the mapping $\alpha$ wherein $\alpha\beta=1_T$, we have a similar situation. Using the same line of logic, we see that there cannot be $s\in S$ such that $s\ne \beta(t) $. If there were, $\beta$ would cease to be unique. Thus, we have that $\alpha$ is injective. Therefore, $\alpha$ is bijective. $\blacksquare$
A: If $\beta:T\rightarrow S$ exists with $\beta\alpha=id_{S}$ then
we find easily that $\alpha\left(s\right)=\alpha\left(s'\right)$
implies that $s=\beta\alpha\left(s\right)=\beta\alpha\left(s'\right)=s'$
showing that $\alpha$ is injective. 
Conversely let $\alpha:S\rightarrow T$
be injective. If $s_{0}\in S$ then we can construct $\beta:T\rightarrow S$
by sending each element $t=\alpha\left(s\right)\in\alpha\left(S\right)$
to the unique $t\in T$. Elements not contained in $\alpha\left(S\right)$
can be sent to $s_{0}$. Then $\beta\alpha=id_{S}$. Note however
that this does not have to work if $S=\emptyset$. We have the unique empty
map $\alpha:\emptyset\rightarrow T$ wich is vacuously injective.
Only if also $T=\emptyset$ then there exist the empty map $\beta:T=\emptyset\rightarrow\emptyset$
and indeed $\beta\alpha=id_{\emptyset}$. However, if $T\ne\emptyset$ then
no map $\beta:T\rightarrow\emptyset$ exists. 
So the statement: if
$\alpha:S\rightarrow T$ is injective then $\beta\alpha=id_{S}$ for
some $\beta:T\rightarrow S$, is true under the extra condition that
$S\ne\emptyset\vee T=\emptyset$. 
If $S$ is a singleton then automatically
$\beta$ is unique. However, if moreover $T$ is not a singleton then
$\alpha$ is not a bijection. So uniqueness of $\beta$ does not imply
that $\alpha$ is bijective. For this we need the extra condition that $S$ is no singleton or $T$ is a singleton.

If $\alpha\beta=id_{T}$ then $t=\alpha\left(\beta\left(t\right)\right)$
for each $t\in T$ showing immediately that $\alpha$ is surjective.
Conversely let $\alpha:S\rightarrow T$ be surjective. Then we construct
$\beta:T\rightarrow S$ by sending each element $t$ by one of an
elements $s\in S$ that suffices $\alpha\left(s\right)=t$. Then automatically
$\alpha\beta=id_{T}$. Note that this equation implies that $\beta(t)$ belongs to fibre $\alpha^{-1}\left(\left\{ t\right\} \right)$, so that is necessary.
If this $\beta$ with $\alpha\beta=id_{T}$ is unique
then for each $t\in T$ the fibre $\alpha^{-1}\left(\left\{ t\right\} \right)$
contains exactly one element. This tells us that the surjective $\alpha$
is also injective, hence bijective.

In category Sets every epimorphism (surjection) is a retraction (second case), but not every monomorphism (injection) is a section (first case). Exceptions are the elements in $\mathbf{Sets}\left(\emptyset,T\right)$ where $T\ne\emptyset$. They are injective but are not sections.
I have been working with sets. If you are working in groups, rings, et cetera then the underlying sets will not be empty.
