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In category theory it is often said that certain operations are defined "pointwise". For example, limits in a functor category $[\mathcal{C},\mathcal{D}]$ can be defined "pointwise" (if $\mathcal{D}$ has these limits). Similarly, if $\mathcal{D}$ carries a monoidal structure, we can define a monoidal structure on $[\mathcal{C},\mathcal{D}]$ "pointwise". By this one actually means "objectwise". It refers to the objects of $\mathcal{C}$, which are usually not called points, even less in specific examples of categories. So why does one still use this terminology "pointwise"? And what do you think of using "objectwise" in a paper instead, will the reader still immediately understand the meaning? And don't we need a terminology which can be used without quotation marks? Is there one which is established in this context? Which one do you recommend?

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    $\begingroup$ Google objectwise "category theory" to find plenty of examples of people using the term in print — both in category theory itself, and in other fields making use of it. The first five results I get are from Emily Riehl, Bertrand Toën, Georg Biedermann, P.J. Hilton, and Steve Awodey — plenty established names. $\endgroup$ Commented Apr 2, 2017 at 10:13
  • $\begingroup$ "pointwise" could lead to other misunderstandings: some (e.g. closed monoidal) categories $\mathcal C$ have a unit object $E$ (like $\Bbbk$ in the category of vector spaces), and then one can model "points" in an object $O$ by maps $E\to O$, even if one does not have an "underlying set" functor $\mathcal C\to\mathcal Set$. $\endgroup$ Commented Apr 2, 2017 at 11:16

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As Peter said in a comment, "objectwise" is quite common and would probably be understood even without quotation marks, although it never hurts to define a term the first time you use it. I expect the reason that some people use "pointwise" is by analogy to functions between sets: the pointwise sum of $f,g:\mathbb{R}\to\mathbb{R}$ is $(f+g)(x)=f(x)+g(x)$, so the "pointwise tensor product" of $F,G:\mathcal{C}\to\mathcal{V}$ is $(F\otimes G)(C) = F(C) \otimes G(C)$.

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