1
$\begingroup$

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 $(x \Rightarrow G)$.

$\endgroup$
  • $\begingroup$ 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
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.