3
$\begingroup$

I was told that there is only one morphism in $Mon$ category for this object $(\mathbb{N},\times)$. But why?

I think that we can write every natural number as the product of a certain set of prime numbers. So we can say that the set of primes forms the basis of this monoid $(\mathbb{N},\times)$, that is, linear combinations with positive integer coefficients (powers) make up the entire set of supports for the structure of this monoid: for any $n\in\mathbb{N}$, $n = p_1^{g_1}\times\cdots\times p_i^{g_i}$, where $\forall p_j \in \{p_1,p_2,\cdots\}$ -- basis, and $\forall g_j\in \mathbb{Z_{>0}}$. And we can make some transpositions on the elements of the basis $\{p_1,p_2,\cdots\}$ thereby defining the morphism (where always $1$ going to $1$ by this morphism). So we have infinite number of morphisms, I am right?

$\endgroup$
  • 3
    $\begingroup$ Are you sure they said $(\mathbb{N},\times)$ and not $(\mathbb{N},+)$? $\endgroup$ – Eric Wofsey Dec 16 '18 at 2:23
  • $\begingroup$ @EricWofsey ye, multiplication $\endgroup$ – Just do it Dec 16 '18 at 2:32
  • 2
    $\begingroup$ Yes an infinite number of morphisms $\endgroup$ – BananaCats Category Theory App Dec 16 '18 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.