6
$\begingroup$

Is there a standard name for taking a homomorphism from the free object over an algebraic structure? Roughly speaking, this should amount to evaluation of any element of the free object under the homomorphism.

For example, take the free commutative monoid over the set of natural numbers; elements of this monoid are finite multisets of natural numbers. By taking commutative monoid homomorphisms we can express many common operations, such as cardinality (map every generator to 1 and take sum as the monoid operation), sum (take every generator to itself and map the monoid operation to sum of natural numbers), product (take product of natural numbers rather than sum) etc.

Since these operations are all abelian monoid operations, we can expect a sensible result which will not depend on our multiset being non-empty or listing its elements in a specific order.

Does this idea generalize to other algebraic structures? Any known references?

$\endgroup$

3 Answers 3

3
$\begingroup$

The nLab has proposed the name "adjunct", which seems OK to me. So, for example, you could say something like "let $F(S)$ be the free group on the set $S$, let $G$ be a group, let $f:S\to G$ be a set map, and let $g:F(S)\to G$ denote the adjunct of $f$." I think it would be even a bit better to say "left adjunct" instead of just "adjunct".

I think Mac Lane might have used the term "transpose".

$\endgroup$
3
  • 2
    $\begingroup$ Mac Lane used "left adjunct" in Theorem IV.1 (ii) ("...while each $g:x\to Ga$ has left adjunct [...] $Fx\to a$") $\endgroup$
    – user2734
    Commented Jan 25, 2010 at 22:50
  • $\begingroup$ You mean adjoint functors, probably. $\endgroup$ Commented Mar 6, 2016 at 2:52
  • 1
    $\begingroup$ @VladPatryshev The functors are said to be adjoint, and applying the adjunction to a morphism is said to yield its adjunct (left/right depending) $\endgroup$
    – ebrahim
    Commented Mar 6, 2016 at 2:55
5
$\begingroup$

This is almost identical to what a computer scientist might call a catamorphism. In this case, the algebraic structure is an initial F-algebra rather than a free commutative monoid. But just as you describe, operations like cardinality, sum and product of lists (as opposed to multisets) can all be described as catamorphisms. It's not an exact correspondence because commutative monoids don't form initial F-algebras in Set. (Though free monoids in general do, and I suspect that free commutative monoids form F-algebras in the category of monoids by Eckmann-Hilton.)

Mathematicians tend to call these kinds of things "canonical" maps.

$\endgroup$
0
$\begingroup$

Given an monad $F$ in a category $C$, we can always build an algebra $FFX \rightarrow FX$ over $FX$.

For a morphism $f:X \rightarrow A$, where $A$ is an algebra over $F$, that is, a morphism $FA \rightarrow A$, with all the right properties, we can naturally extend $f$ to a morphism between two algebras, $f': FX \rightarrow A$. This morphism, from a free algebra to an algebra, is called catamorphism.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .