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Let $\mathcal{C}$ be a complete category and suppose $f: X \to X$ is an endomorphism in $\mathcal{C}$. Associated to $f$ is an inverse system, $$X_\bullet: \dots \to X \to X \to X \to X,$$ where every arrow is $f$, and we can form a map of inverse systems $f_\bullet: X_\bullet \to X_\bullet$, again just via $f$. Taking inverse limits, we arrive at $\hat{f}: \hat{X} \to \hat{X}$.

In any concrete category where inverse limits can be represented as sets of coherent sequences, it's easy to see that $\hat{f}$ is an epimorphism. Indeed, given a coherent sequence $(x_0, x_1, \dots)$, we have that $$\hat{f}(x_1,x_2,\dots) = (f(x_1),f(x_2),\dots) = (x_0,x_1,\dots).$$

Is it always true that $\hat{f}$ is an epi? If not, what counter-examples exist? What kind of conditions could be imposed on $\mathcal{C}$ to ensure $\hat{f}$ is an epi?

It seems like this holds for a large class of categories (e.g. all complete abelian categories, by Freyd–Mitchell), so I'm curious to see just how general it really is.

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In fact, you can prove that in the category of sets this $\hat{f}$ is even an isomorphism : indeed, in that category $\hat{X}$ can be represented as the set of coherent sequences you mention, and you have a map $$\gamma:\hat{X}\to \hat{X}:(x_0,x_1,x_2,\dots)\mapsto (x_1,x_2,\dots).$$ You have proved already that $\hat{f}\gamma=id_{\hat{X}}$, and it's easy to see that $\gamma \hat{f}=id_{\hat{X}}$ as well.

Now using this fact, you can prove that in fact $\hat{f}$ must be an isomorphism in any complete category. Indeed, by the Yoneda lemma it is enough to prove that $\hat{f}^* : Hom_\mathcal{C}(\_, \hat{X})\to Hom_\mathcal{C}(\_, \hat{X})$ is a natural isomorphism; but for this it is enough to prove that $\hat{f}^* : Hom_\mathcal{C}(Z, \hat{X})\to Hom_\mathcal{C}(Z, \hat{X})$ is an isomorphism for all $Z$.

But the universal property of the limit tells you that for all $Z$, $Hom_\mathcal{C}(Z, \hat{X})$ must be (naturally isomorphic to) the limit of $$\dots Hom_\mathcal{C}(Z, X)\to Hom_\mathcal{C}(Z, X)\to Hom_\mathcal{C}(Z, X)$$ (where every map is $f^*$) in the category of sets, and you can prove that $\hat{f}^*=\widehat{f^*}$ which is thus an isomorphism.

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  • $\begingroup$ I presume in your final indented line you should write $X$ not $\hat{X}$ in the homs? But thanks for the enlightening answer. $\endgroup$
    – Mr. Chip
    Nov 16, 2017 at 23:19
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    $\begingroup$ Here is a fun example: let $X = S^1$ in the category of compact Hausdorff abelian groups, and let $f : X \to X$ be given by $z \mapsto z^p$. Then $\hat{X}$ is a group called the $p$-adic solenoid; it can be thought of as the Pontryagin dual of $\mathbb{Z}[1/p]$ (which arises via the dual construction of a colimit rather than a limit in abelian groups). The identification of the Pontryagin dual gives some indication of what this process has to do with attempting to invert $f$. $\endgroup$ Nov 17, 2017 at 9:43
  • $\begingroup$ The map $x \mapsto x^p$ (and the process of "perfecting" an $\mathbb{F}_p$-algebra it's defined on) was indeed what inspired this question! $\endgroup$
    – Mr. Chip
    Nov 17, 2017 at 10:24

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