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Let $f: M \to N$ be a mono in $\mathsf{Mon}$. I want to prove that the underlying function $U(f): U(M) \to U(N)$ is injective.

How can this be done ? Thanks in advance.

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  • $\begingroup$ By Mon you mean the category of monoids..? $\endgroup$ – Piotr Pstrągowski Sep 30 '13 at 16:43
  • $\begingroup$ Yes, sorry for not mentioning it. $\endgroup$ – user42761 Sep 30 '13 at 16:44
  • $\begingroup$ This is true because $U$ is a right adjoint. $\endgroup$ – Zhen Lin Sep 30 '13 at 16:44
  • $\begingroup$ I don't know what a right adjoint is. Could you maybe explain this ? $\endgroup$ – user42761 Sep 30 '13 at 16:45
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It's clear that injective maps are monic, so let's do the converse. Let $f: M \rightarrow N$ be monic in Mon and let $m_{1}, m_{2} \in M$ satisfy $f(m_{1}) = f(m_{2})$. Consider the functions $\phi_{1}, \phi_{2}: \mathbb{N} \rightarrow M$ determined by $\phi _{i}(1) = m_{i}$. Since $f$ is monic and we have $f \phi_{1} = f \phi_{2}$, we must have $\phi _{1} = \phi _{2}$. Hence $m_{1} = m_{2}$.

This is in general true for all algebraic categories that admit a "free object" construction, ie. there exists a left adjoint to the forgetful functor. Here it comes into play since $\mathbb{N}$ is a free monoid on one element set.

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  • $\begingroup$ Oh ok. This looks like the UMP of the monoid with just one element. Thanks. $\endgroup$ – user42761 Sep 30 '13 at 17:22

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