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For an object $X$ of a category, $h_X$ is the contravariant functor represented by $X$, i.e. $h_X = Hom(-,X)$.

Question a) Who invented this notation? (My guess: Grothendieck)

b) Is there a special reason why the letter $h$ was chosen? Is it in an abbreviation for "homomorphism"?

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    $\begingroup$ I dunno, but I think the more interesting case is $h^X:=Hom(X,-)$, where it's a superscript because it is contravariant in $X$. $\endgroup$ Commented Feb 2, 2011 at 9:49
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    $\begingroup$ @Harry: Offtopic. $\endgroup$ Commented Feb 2, 2011 at 10:54
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    $\begingroup$ Hey, don't be rude about it! If my comment is off topic, then your entire question is most certainly so. $\endgroup$ Commented Feb 2, 2011 at 17:22

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Let me answer the questions in order.

a) It was invented by Grothendieck, see EGA I, Springer edition, especially chapter 0, discussion of representable functors.

b) Quite possibly is a shortcut for $Hom$. Sometimes the letter $y$ is used (for Yoneda). The trouble is when you are considering the representable functor defined over several categories, e.g. a category and a subcategory.

Further evidence on a) The notation is already on SGA 3 and 4. There are several exposés by Grothendieck in Henri Cartan's seminar from 1960/61 in which he explains his point of view of Teichmüller's space through representable functors in the analytical category and he uses the notation $h_X$.

I an not aware of anyone else using these ideas at that time. Cartan's seminar is available at numdam:

http://www.numdam.org/article/SHC_1960-1961__13_1_A7_0.pdf

See also Bourbaki seminar, exposé 195 (February 1960)

http://www.numdam.org/article/SB_1958-1960__5__369_0.pdf

Bonus: If you, instead of considering contravariant functors $\mathrm{Sch}^{o} \to \mathrm{Set}$, use covariant functors $\mathrm{Aff} \to \mathrm{Set}$ the notation used in EGA is $h_X^{o}$. Perhaps the reason is that Yoneda's map is contravariant in this case.

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  • $\begingroup$ Of course the notation $h_X$ is used extensively in EGA, but where can you find evidence that that it is Grothendieck's invention? $\endgroup$ Commented Feb 2, 2011 at 10:53
  • $\begingroup$ I add evidence to this question in an edit. $\endgroup$
    – Leo Alonso
    Commented Mar 24, 2020 at 21:27

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