In their book "Non-Archimedean analysis", when BGR refer to an epimorphism in the context of $k$-Banach algebras do they actually require (or does it follow that) such maps are surjections? For example, after their definition of $k$-affinoid algebras, they seem to use Banach's Open mapping theorem to deduce such algebras are quotients of Tate algebras, however that would require surjectivity.
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$\begingroup$ A non-surjective map of Banach algebras may well have dense image, and in that case it sstill is an epi. An examplee is the inclusion of $C^1(M)$ in $C^0(M)$ for $M$ a closed manifold. $\endgroup$– Mariano Suárez-ÁlvarezCommented Sep 10, 2013 at 6:06
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$\begingroup$ Thanks for the answer&comment. This clears up some confusion. How do you think the authors intended us to read "epimorphism" as surjection? It doesn't seem like a simple abuse of language, since the authors use both words in the context of rings. $\endgroup$– LMNCommented Sep 10, 2013 at 12:54
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$\begingroup$ It is quite common, especially in areas far away from category theory, that "epimorphism" actually means "surjective homomorphism". In some sense this is OK since in many interesting categories epimorphisms are just awful and useless, but surjective homomorphisms are easy to describe and useful. $\endgroup$– Martin BrandenburgCommented Sep 10, 2013 at 22:16
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The authors mean a surjective homomorphism, "epimorphism" is too weak. An affinoid $k$-algebra is defined to be a Banach $k$-algebra of the form $k\langle x_1,\dotsc,x_n \rangle/\mathfrak{a}$ for some closed ideal $\mathfrak{a}$. This is analogous to affine $k$-algebras which have the form $k[x_1,\dotsc,x_n]/\mathfrak{a}$. So basically polynomials are replaced by strictly convergent power series.
answered Sep 10, 2013 at 9:44