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Aviles and Koszmider give an example of a co-Hopfian Banach space here. So, this space $X$ has the property that any continuous linear one-to-one operator on $X$ is an isomorphism in the category of Banach spaces and continuous linear transformations.

Brian M. Scott gave answer here using the fact that shift operators in sequence spaces are one-to-one but not isomorphisms . An example of such an operator is the map sending the sequence $\{x_1, x_2, \ldots \}$ to the sequence $\{ 0, x_1, x_2, \ldots\}.$

Do we know any examples of Hopfian Banach spaces, those $X$ for which any continuous onto map $X \rightarrow X$ is an isomorphism?

Chris Leary
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    Wild guess: the dual of Aviles and Koszmider's space? – Ruy Jan 11 '21 at 19:49
  • @Ruy - Interesting idea. If I knew more functional analysis, I could probably say something intelligent in response. – Chris Leary Jan 11 '21 at 20:22
  • @Ruy - After thinking about this for a bit, it seems that the problem has a relation to Matlis duality, which behaves well with respect to the Hopfian and co-Hopfian properties as noted by user Mohan on MSE. Your "wild guess" may not be so wild. I'll keep on it. – Chris Leary Jan 12 '21 at 18:54
  • @Ruy - I think you are correct. Since I need to work only with one-to-one and onto conditions, Martin's answer in https://math.stackexchange.com/questions/300435/is-duality-an-exact-functor-on-banach-spaces-or-hilbert-spaces shows that taking dual is an exact functor and Mohan's argument will work. In the same question, Nate Eldredge shows that in the categorical sense, where onto, for instance, is replaced by epimorphism (that is, has dense range), then things no longer work. thanks for steering me in what seems to be the correct direction. – Chris Leary Jan 12 '21 at 21:17
  • Good to hear the good news. Also to know that wild guesses are sometimes useful $\ddot\smile$ – Ruy Jan 12 '21 at 21:35

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