"Perfecting" an endomorphism in a category 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.
 A: 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.
