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

![naturality](http://i64.tinypic.com/vglx7n.jpg)

conmutes?