I'm lagging a bit, but a question: I'm working to see the general idea that the German-Italian business exemplify, and seeking for a more systematic grip.

Would it be correct to say that in our example \\(Lan_G\\) sends sets to identities? (or better, sends the things that correspond to sets in \\(\mathbf{Set^1}\\) to an identity on that set?). Or better, since left adjoint functors are unique only up to natural isomorphism, that a functor obeying what I say is *a* left adjoint to \\(- \circ G\\)?

In more detail, be \\(Ob(\mathbf{1})=\\{\star\\}\\), and \\(Ob(\mathbf{2})=\\{0, 1\\}\\), \\(Mor(\mathbf{2})=\\{Id_0, Id_1, i:0 \to 1\\}\\) and \\(G(\star)=0\\).

Let \\(S_1,S_2:\mathbf{1} \to \mathbf{Set}\\)

\\(Lan_G:\mathbf{Set}^\mathbf{1} \to \mathbf{Set}^\mathbf{2}\\) acts on objects and morphisms via functions

\[ Lan_G^{Ob}: Ob(\mathbf{Set}^{\mathbf{1}}) \to Ob(\mathbf{Set}^{\mathbf{2}}) \]
\[ Lan_G^{Mor}: Mor(\mathbf{Set}^{\mathbf{1}}) \to Mor(\mathbf{Set}^{\mathbf{2}}) \]

So if I submit a functor \\(S_1\\) to \\(Lan_G\\), I obtain a functor of \\(\mathbf{Set}^{\mathbf{2}}\\).

I'am asking if it's OK to say that \\(Lan_G\\) sends \\(S_1\\) (and hence in a way its associated set \\(S_1(\star)\\)) to the identity on it, \\(Id_{S_1(\star)}\\).

Yet more detailedly, can I say that \\(Lan_G\\) on objects does this to \\(S_1\\):

\[ Lan_G^{Ob}(S_1)(0) = S_1(\star) \]
\[ Lan_G^{Ob}(S_1)(1) = S_1(\star) \]
\[ Lan_G^{Ob}(S_1)(i) = Id_{S_1(\star)} \]

and similarly for \\(S_2\\)

\[ Lan_G^{Ob}(S_2)(0) = S_2(\star) \]
\[ Lan_G^{Ob}(S_2)(1) = S_2(\star) \]
\[ Lan_G^{Ob}(S_2)(i) = Id_{S_2(\star)} \]

...and on arrows, for \\(\mu:S_1 \Rightarrow S_2\\), so \\(\mu \in hom_{\mathbf{Set}^\mathbf{1}}(S_1, S_2\\)) with \\(\mu_\star:S_1(\star) \to S_2(\star)\\), it assigns

\[ Lan_G^{Mor}(\mu):=\tau \]

with \\(\tau = \\{\tau_0 := \mu_\star, \tau_1 := \mu_\star\\}\\) ("duplicates arrows") so