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.
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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 $$ X \overset{f}{\underset{h}{\rightleftarrows}} Y \overset{g_*}{\underset{d_*}{\rightleftarrows}} Z$$ |
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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. |
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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. |
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