If there exists no injection $X \rightarrow Y$ does that mean that there exists a surjection $Y \rightarrow X$ ?
If there exists no surjection $X \rightarrow Y$ does that mean that there exists an injection $Y \rightarrow X$ ?
In other words, are two sets always comparable by size ?
EDIT: you've made a mistake in how you phrase comparability of sets. What you've written is false: take e.g. $X=\{1,2\}, Y=\{1\}$ for the first version and $X=\{1\}, Y=\{1, 2\}$ for the second version.
The right statement of comparability is:
For any sets $X, Y$, either there is an injection from $X$ to $Y$ or there is an injection from $Y$ to $X$.
And (at least, in the presence of choice), we can also use surjections instead: for any sets $X, Y$, there is an injection from $X$ to $Y$ iff there is surjection from $Y$ to $X$. This means we can equivalently restate comparability as:
For any sets $X, Y$, either there is a surjection from $X$ to $Y$ or there is a surjection from $Y$ to $X$.
But in general, in the absence of choice we can't alternate between injections and surjections this way: the axiom of choice is in fact equivalent to the statement "for any sets $X, Y$, there is an injection from $X$ to $Y$ iff there is surjection from $Y$ to $X$."
This is exactly what the axiom of choice is all about! (Or at least, it's one of the things; AC is prolific.)
The comparability of all sets with respect to cardinality is a consequence of the axiom of choice. In fact, it's equivalent to it (over the usual axioms for set theory without choice, ZF): ZF proves that the axiom of choice holds iff cardinalities are always comparable. Meanwhile, there are reasonable set theories (like ZF+AD) in which the answer is "no."
In general, choice is assumed unless otherwise stated - so the context-free answer to your question would be "yes."
There are a couple different ways to prove trichotomy from choice:
Use Zorn's lemma, taking as the relevant poset the set of all partial injections from one set to the other, and think about what a maximal element is.
Use the well-ordering theorem, and show that any two well-orderings are comparable.
To prove choice from trichotomy is actually quite simple, and uses a neat fact: that for any set $X$, there is a well-orderable set $Y$ which does not inject into $X$. (This is Hartogs' theorem.) Now note that trichotomy then implies that every set injects into a well-orderable set, hence is well-orderable (why?), so we have the well-ordering theorem; and proving choice from the well-ordering theorem is easy.
