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.


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.

  • $\begingroup$ 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 '17 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 $f$. $\endgroup$ – Qiaochu Yuan Nov 17 '17 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 '17 at 10:24

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