# 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?

• Injective iff left inverse is actually not true; it doesn't hold for the function from the empty set to any inhabited (= non-empty) set. Still, it's easier to characterize them in ZF. Apr 15, 2020 at 1:38
• On the other hand, note that injectivity is characterized by a $\forall$ statement, whereas surjectivity is characterized by a $\forall\exists$; that’s already a significant difference. Caveat: kernels work for groups, rings, vector spaces, modules. But they don’t work, e.g., for semigroups. Apr 15, 2020 at 4:49
• Btw: for a completely symmetric/dual characterization of surjective and injective functions that holds in ZF, a function is injective iff it is left cancellable, and a function is surjective iff it is right cancellable. Apr 15, 2020 at 4:50
• @Omnomnomnom: "right cancellable" means that if $ac = bc$, then $a=b$. If $c$ has a right inverse, it must be right cancellable, but the converse isn't true. For example, in $\mathbb{Z}$, $a\cdot 3 = b\cdot 3$ implies $a=b$ even though $\frac{1}{3}\notin\mathbb{Z}$. Apr 15, 2020 at 5:00
• Another distinction is that surjectivity only makes sense when you think of functions as triples (domain, codomain, set of pairs), whereas injectivity also makes sense when you think of functions as simply sets of ordered pairs satisfying $(a,b),(a,b’)\in f\implies b=b’$, making “injectivity” a slightly more “primitive” notion than surjectivity, despite the general duality. Apr 15, 2020 at 5:39

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.