Are injectivity and surjectivity dual? Are injectivity and surjectivity dual in some sense? Their set-theoretic definitions are quite different. In particular, the injectivity is a property of a function's graph, while surjectivity is a relationship between the range of the function and its codomain.
 A: There is an important sense in which they are not strictly dual.
Let X is a non-empty set.
Then f: X --> Y is injective if and only if it has a left inverse. The proof does not require the Axiom of Choice.
But it is not true that f:X --->Y is surjective if and only it has a right inverse, unless you invoke the Axiom of Choice. In fact, this is logically equivalent to the Axiom of Choice.
A: Let $K$ be a field, let $V$ and $W$ be $K$-vector spaces, and let $f: V \rightarrow W$ be a $K$-linear map.  Let $V^{\vee} = \operatorname{Hom}(V,K)$ and $W^{\vee} = \operatorname{Hom}(W,K)$ be the dual spaces.  Then $f$ induces a map 
$f^{\vee}: W^{\vee} \rightarrow V^{\vee}$, by $\ell \in \operatorname{Hom}(W,K) \mapsto (v \in V \mapsto \ell(f(v)))$.  
1) $f$ is surjective iff $f^{\vee}$ is injective.
Suppose $f$ is surjective and that there is $\ell \in W^{\vee}$ with $f^{\vee}(\ell) = 0$.  That is, for all $v \in V$, $\ell(f(v)) = 0$.  Since $f$ is surjective this means that $\ell(w) = 0$ for all $w \in W$ and thus $\ell = 0$.  
Suppose $f$ is not surjective, and let $w \in W \setminus f(V)$.  Then there is a linear functional $\ell$ on $W$ which vanishes identically on $f(V)$ but not at $w$.  Thus $\ell$ is not $0$ but $f^{\vee}(\ell)$ is, so $f^{\vee}$ is not injective.
2) $f$ is injective iff $f^{\vee}$ is surjective. 
Suppose $f$ is injective.  Then we may view $V$ as a subspace of $W$, and the map $f^{\vee}$ is just restriction of linear functionals to a subspace.  This is clearly surjective, since any linear map on a subspace can be extended to a linear map on the ambient space.
Suppose $f$ is not injective: let $0 \neq v \in V$ be such that $f(v) = 0$.  Then no linear functional on $V$ with $L(v) \neq 0$ lies in the image of $f^{\vee}$, so $f^{\vee}$ is not surjective.
A: Yes, in some sense.  
Surjections and injections are "categorical duals" (in $\mathbf{Set}$). The simplest way to see a manifestation in the duality is the following: let $X,Y,Z$ be sets. If $f:X\to Y$ is surjective, then for any $g_1,g_2:Y\to Z$, 
$$g_1\circ f = g_2 \circ f \iff g_1 = g_2$$
Whereas if $h: Y\to X$ is injective, then for any $d_1,d_2: Z\to Y$ 
$$ h\circ d_1 = h\circ d_2 \iff d_1 = d_2 $$
The duality is clearer if you draw the diagram
$$ X \overset{f}{\underset{h}{\rightleftarrows}} Y \overset{g_*}{\underset{d_*}{\rightleftarrows}} Z$$
