The following is a definition given in my lecture notes that I'm not sure is standard, so I'll write it out: Given a functor $G: \mathcal{D} \to \mathcal{C}$ and an object $x$ in $\mathcal{C}$, we have the category $(x \Rightarrow \mathcal{C})$ where:

  • Objects are pairs $(y, f)$ where $y \in \mathcal{D}$ and $f: x \to G(y)$;
  • Morphisms $(y_1, f_1) \to (y_2, f_2)$ are $\mathcal{D}$-morphisms $\alpha: y_1 \to y_2$ such that $G(\alpha) \circ f_1 = f_2$.

I'm trying to prove that $F \dashv G$ if and only if $(Fx, \eta_x)$ is initial in $(x \Rightarrow G)$ for some $\eta: \mathrm{id}_{\mathcal{C}} \Rightarrow GF$.

I can see that any morphism $(Fx, \eta_x) \to (y, f)$ must be unique, but I am struggling to see why there should be such a morphism for every $(y, f)$ in $(x \Rightarrow G)$.

  • $\begingroup$ what do you mean by $F\dashv G$? $\endgroup$ – Felipe Pérez May 10 '16 at 14:09
  • $\begingroup$ I mean F is left-adjoint to G, where F is a functor $\mathcal{C} \to \mathcal{D}$ $\endgroup$ – Alex McKenzie May 10 '16 at 14:09
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    $\begingroup$ Also, if you need it, I find really clear and linear the explanation in Leinster, "Basic category theory" (chapter 2 section 3, theorem 2.3.6 pag. 61 is what you are searching for). $\endgroup$ – any_one Jan 14 '17 at 13:22

Given an object $(y,f)$, you have $f : x \to G(y)$. The adjoint transpose of $f$ is $$\alpha := \varepsilon_y \circ F(f) : F(x) \to y$$ Moreover $$\begin{align} G(\alpha) \circ \eta_x &= G(\varepsilon_y) \circ GF(f) \circ \eta_x && \text{by definition of } \alpha \\ &= G(\varepsilon_y) \circ \eta_{G(y)} \circ f && \text{by naturality of } \eta \\ &= f && \text{by the triangular identities} \end{align}$$

So $\alpha$ defines a morphism $(Fx,\eta_x) \to (y,f)$ in the comma category.


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