Lemma 4.24.2 (0A8B)—The Stacks project

Lemma 4.24.2. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. If for each $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ the functor $x \mapsto \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(u(x), y)$ is representable, then $u$ has a right adjoint.

Proof. For each $y$ choose an object $v(y)$ and an isomorphism $\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(-, v(y)) \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(u(-), y)$ of functors. By Yoneda's lemma (Lemma 4.3.5) for any morphism $g : y \to y'$ the transformation of functors

$\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(-, v(y)) \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(u(-), y) \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(u(-), y') \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(-, v(y'))$

corresponds to a unique morphism $v(g) : v(y) \to v(y')$ . We omit the verification that $v$ is a functor and that it is right adjoint to $u$ . $\square$

Comments (3)

Comment #7932 by olof on December 20, 2022 at 02:44

"For each y choose an object v(y)"Why can we 'choose' v(y) instead of 'there exists a U in category C'?The definition of representable functor says that 'exists'.

Comment #7933 by olof on December 20, 2022 at 03:57

And,if that isomorphism is true,then proof is no need

Comment #7934 by olof on December 20, 2022 at 04:06

ok i understanded

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