I don't understand this definition:

We say that a functor $L:{\cal{C}}\rightarrow {\cal{D}}$ is left adjoint for a functor $R:{\cal{D}}\rightarrow {\cal{C}}$ iff there is a natural isomorphisms $$\theta_{C,D}:\text{Hom}_{\cal{D}}(LC,D)\rightarrow \text{Hom}_{\cal{C}}(C,RD)$$ for all $C\in{\cal{C}},D\in{\cal{D}}$

What does this mean?

Is it in some sense the natural transformation between two functors or are we just looking for an isomorphism between the two sets (I wouldn't know what that means: just a bijection?)

EDIT:

I got the definition from ಠ_ಠ

But know to make sense of it $$F:=\operatorname{Hom}(L(-), -): \mathsf{C}^{\mathrm{op}} \times \mathsf{D} \to \mathsf{Sets}$$ sends $(C,D)$ to $\operatorname{Hom}(L(C),D)$ and acts on functions as follows:

I take $$(f,g):(C',D)\rightarrow (C,D')$$ Then $$F(f,g):F(C',D)\rightarrow F(C,D'):h\mapsto g\circ h\circ Lf$$

But how do I define $$G:= \text{Hom}(-,R(-))$$ on functions? The arrows go the wrong way.

• $\matrix{LC' & \to &D \\ \uparrow {\scriptstyle Lf} && {\scriptstyle g} \downarrow \\ LC && D'}$ – Berci Dec 3 '17 at 0:14
• Makes sense. How do I define $Hom(-,R(-))$ on $(f,g)$ the arrows go the wrong way. – tomak Dec 18 '17 at 9:57

$\operatorname{Hom}(L(-), -): \mathsf{C}^{\mathrm{op}} \times \mathsf{D} \to \mathsf{Sets}$ and $\operatorname{Hom}(-, R(-)): \mathsf{C}^{\mathrm{op}} \times \mathsf{D} \to \mathsf{Sets}$ are both functors, and the definition states that $L$ is left adjoint to $R$ if these aforementioned Hom functors are naturally isomorphic. Just draw out the naturality squares for naturality in each variable; they're printed on page 184 of Awodey's Category Theory text if you get stuck. And yes, an isomorphism of sets is a bijection.
Edit: to answer the second part of your question, $\mathrm{Hom}(f,g)=f^* g_*$. Where the lower star indicates post-composition, and the upper star indicates precomposition.
• Why $C^{op}$? Shouldn't it be Hom$(L(-),-):{\cal{C}}\times{\cal{D}}\rightarrow$ Set? – tomak Dec 2 '17 at 11:22
• better :) but still why won't a simple $C$ instead of $C^{op}$ work? – tomak Dec 2 '17 at 11:26