Why is surjectivity "harder" than injectivity? Injectivity and surjectivity are very intimately related, however, in any particular structure, one of them tends to be a much "harder" property - just looking at basic set theory, we have that a function is injective iff it has a left inverse, and surjective iff it has a right inverse. But immediately something crops up: the former is a harmless statement provable in ZF, but the latter is equivalent to the axiom of choice. 
Going on to basic algebra, we see that kernels  are totally related to the injectiveness of a function, and studying them tend to be much easier, while cokernels tend to bring in no new information (and at least, at the lower level, seem to be just a fancy but useless substitute for studying  surjectivity). We also see that theorems about surjectivity tend to be more important (off the top of my head, the isomorphism extension theorem in field theory). Surjectivity questions tend to have to be answered constructively, which in my experience, is generally hard.
That is not to say that the opposite isn't true - from memory, projective resolutions were significantly easier than injective ones.
My question is: is there a "deep" reason as to why one of the two tend to be much "harder" than the other?
 A: I’d try to explain this as follows: we are in the habit of understanding a set $S$ in terms of the maps from a singleton $*$ to $S$. This breaks the symmetry dramatically between “epimorphism” and “monomorphism,” the left- and right-cancellations conditions discussed in the comments. 
Indeed, if you know $f:S\to T$ is a monomorphism, then you know immediately from the definition that $f$ does not identify any two distinct points $x,y:*\to S$. But if you know $f$ is an epimorphism, it’s much less obvious what this means in terms of maps from the point into $S$ and $T$. You have to know a lot more about the structure of sets to say that, if there were any point of $T$ not in the image of $f$, then one could construct two unequal maps out of $T$ equalized by $f$. And these maps out of $T$ wouldn’t most naturally be into any nice fixed set like $*$! At best, you could use maps $T\to \{0,1\}$.
In fact, if you know enough about $\{0,1\}$, you know you can characterize the epis as precisely the maps $f$ which induce a mono on powersets, $f^*:\{0,1\}^T\to \{0,1\}^S$. So one might measure the difference in difficulty of injective versus surjective maps by considering how much more complex the powerset is than the mere set. For instance, the powerset construction isn’t even available in most other categories, though $\mathbb k$ serves the same role among finite dimensional vector spaces.
