I am interested whether the following construction naturally appearing in Commutative Algebra has some know and acceped name.

Given a commutative monoid $(M,+)$ and a set $X$, consider the family $F(X,M)$ of functions $\varphi:X\to M$ that have finite support $supp(\varphi):=\{x\in X:\varphi(x)\ne 0\}$ where $0$ is the neutral element of $M$ with respect to the commuative operation $+$.

The set $F(X,M)$ has an obvious structure of commutative monoid (actually a submonoid of the power $M^X$).

Any function $f:X\to Y$ between sets induces a monoid homomorphism $Ff:F(X,M)\to F(Y,M)$ that assigns to each $\varphi\in F(X,M)$ the function $\psi:Y\to M$, $\psi:y\mapsto \sum_{x\in f^{-1}(y)}\varphi(x)$ (the latter sum is well-defined since it contains only finitely many non-zero terms).

The construction $F(X,M)$ determines a functor $F:\mathbf{Set}\to \mathbf{Mon}$ from the category $\mathbf{Set}$ of sets to the category $\mathbf{Mon}$ of commutative monoids.

If am interested if the functor $F$ has some known reserved name.

Remark. For some special monoids $M$ the functor $F$ is well-known in Algebra. For example,

$\bullet$ for the group $\mathbb Z$ of integers, the monoid $F(X,\mathbb Z)$ can be identified with the free Abelian group of $X$;

$\bullet$ for the 2-element cyclic group $C_2$, the the monoid $F(X,C_2)$ can be identified with the free Boolean group of $X$.

$\bullet$ for the n-element cyclic group $C_n$, the the monoid $F(X,C_n)$ can be identified with the free Abelian group of $X$ in the variety of Abelian groups satisfying the identity $x^n=1$;

$\bullet$ for the 2-element semilattice $2=\{0,1\}$ with operation $\max$, the monoid $F(X,2)$ can be identified with the free semilattice with unit over $X$.

Added in Edit. I see that besides downvotes no good name for the functor $F$ was suggested. I perfectly understand that $F(X,M)$ is the direct sum of $X$ copies of $M$. But this cannot be written as a short name of the functor. Or call it "the functor of $M$-th copower" (by analogy with the "functor of $n$th power" assigning to each $X$ its power $X^n$)? My previous idea was "the functor of $M$-valued finitary functions". What is better or more appropriate? Simply, I should call it somehow in a paper. Thanks for constructive comments.

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    $\begingroup$ It's called the direct sum, $\bigoplus_{x \in X} M$. In this case it's usually denoted $M^{(X)}$. $\endgroup$ – Najib Idrissi Feb 2 at 10:06
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    $\begingroup$ It's also the coproduct in the category of commutative monoids, or if you want, the cotensoring by a set... $\endgroup$ – Denis Nardin Feb 2 at 10:54
  • $\begingroup$ @NajibIdrissi The question was about the name. So, call this functor "the functor of direct sum" or "the functor of direct sum?". I already found more-or-less good name: "the functor of $M$-valued finitary functions" but I thought that there existed something more standard. $\endgroup$ – Taras Banakh Feb 2 at 11:30
  • $\begingroup$ @NajibIdrissi Or call this functor "the functor of $M$-th copower" by analogy with the "functor of $n$-th power" that assigns to each $X$ its (finte or infinite) power $X^n$? $\endgroup$ – Taras Banakh Feb 2 at 11:43
  • $\begingroup$ "The functor of $M$-th copower" is more suitable for $(-)^M$, not for $M^{-}$. "The copower functor of $M$" seems better. $\endgroup$ – Oskar Feb 2 at 11:56

Let $X$ be a set and $M$ be a commutative monoid. Then the monoid $F(X,M)$ defined in your question is the direct sum $\bigoplus_{x\in X}M$. Categorically, $\bigoplus_{x\in X}M$ is the coproduct $\coprod_{x\in X}M$ in the category of commutative monoids $\mathbf{CMon}$, or the $X$-th copower of $M$, which also may be denoted by $M^{(X)}$. It was mentioned in the comments.

Note, that copowers are special cases of tensor products in enriched categories (see M.Kelly, "Basic concepts of enriched category theory", p.48 for details), so "copowering" may be also called "tensoring" and we can denote $M^{(X)}=X\otimes M$. This also means that the copowering functor of commutative monoids is actually a bifunctor: $$ \otimes\colon\mathbf{Set}\times\mathbf{CMon}\to\mathbf{CMon}, $$ which sends the pair $(X,M)$ to $X\otimes M=F(X,M)$. See also nlab article, where this functor is called the copowering functor.

As for the functor $F(-,M)\colon\mathbf{Set}\to\mathbf{CMon}$, where the second argument is constant: it may be denoted by $M^{(-)}$ or $-\otimes M$ and called the copower(ing) functor of $M$.

  • $\begingroup$ Thank you for the asnwer. I looked at the nlab article, you mentioned. It too abstract as my may needs. So, many I will retruct to my original "the functor of $M$-valued finitary functions". It is at least self-suggesting. $\endgroup$ – Taras Banakh Feb 2 at 16:19
  • $\begingroup$ Since there are many downvotes (both on my question and your answer, I think it is reasonable for me to remove this question) then your answer also will disappear. Sorry for that. But this will have no influence on your MO score (as I see). $\endgroup$ – Taras Banakh Feb 3 at 14:05
  • $\begingroup$ @TarasBanakh I have no complaints. In any case, I find some parts of your question interesting and worthy of attention. It's up to you whether to remove it. $\endgroup$ – Oskar Feb 3 at 15:31
  • $\begingroup$ It is not so easy to delete question if there was some discussion. So, let it be. Maybe the discussion will be helpful for somebody else (for me it was helpful). $\endgroup$ – Taras Banakh Feb 3 at 16:11

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