I am studying the Yoneda and co-Yoneda lemmas, and in order to understand them well I am trying to develop particular cases. This one is getting me in trouble (it must be easy, but I cannot get it):

Consider the multiplicative monoid $(\mathbb{Z}^*,\cdot)$ of nonzero integers. By multiplication, $\mathbb{Z}_2$ is a right (or left) $\mathbb{Z}^*$-set. According to the co-Yoneda lemma $\mathbb{Z}_2$ is the colimit of the functor :$$\int_{\mathbb{Z}}\mathbb{Z}_2 \xrightarrow {\pi} \mathbb{Z}^* \xrightarrow {Y_*} \widehat{\mathbb{Z}^*}.$$ I want to 'visualize' the shape of that limit.

What I did was to look at $\mathbb{Z}_2$ as a functor $\mathbb{Z}_2\in Ob(\widehat{\mathbb{Z}})$. In this case, $\mathbb{Z}_2(\ast)=\mathbb{Z}_2$ and $\mathbb{Z}_2(n)=\cdot n$. After that, I considered the category of elements, which is essentially $Ob(\int_{\mathbb{Z}}\mathbb{Z}_2)=\mathbb{Z}_2$ and $Mor(\int_{\mathbb{Z}}\mathbb{Z}_2)=\{\overline{0}\xrightarrow{\cdot n}\overline{0}, \overline{1}\xrightarrow{\cdot n}\overline{n}\,|\, n\in \mathbb{Z}^*\}$. With this at hand, I can't see or 'feel' the kind of colimit of $\mathbb{Z^*}$'s (since it is the only representable) $\mathbb{Z}_2$ is.

Here $\mathbb{Z}_2$ are the integers mod 2, $(\mathbb{Z}^*,\cdot)$ is seen as a category with only one object, and $\widehat{\mathbb{Z}^*}=\mathbf{Set}^{(\mathbb{Z}^*)^{op}}$.

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    $\begingroup$ Are all the $\mathbb{Z}$'s supposed to be $\mathbb{Z}^*$? $\endgroup$ – Eric Wofsey Jun 8 '16 at 2:31
  • $\begingroup$ Yes. Thanks for the observation. $\endgroup$ – Luoisv Jun 8 '16 at 5:30

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