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I refer to Rankeya's answer on this question.

Shouldn't $Mor(X,X)$ consist of monomorphisms only, for each morphism to have an inverse?

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  • This does not make sense. In the linked answer, Rankeya defines a category. – Pece Jun 10 '14 at 05:59
  • @Pece- How does Rankeya conclude that every element has an inverse in $Mor(X,X)$. The way I read the answer is that elements of a group can be mapped to elements of $Mor(X,X)$. Hence, $id_G$ maps to the identity mapping, etc. – user1992 Jun 10 '14 at 06:06
  • Rankeya defines $\operatorname{Mor}(X,X)$, which is always a monoid, to be the particular monoid $G$. But $G$ is a group, i.e. a monoid in which every member has an inverse. – Pece Jun 10 '14 at 06:07
  • Actually inverses are also two-sided, so you need more than just being a monomorephism. – user99680 Jun 10 '14 at 06:10
  • @user99680- Yes, but considering we're talking about $Mor(X,X)$, every injective morphism will have a unique two-sided inverse. – user1992 Jun 10 '14 at 06:16
  • Sorry, I now understand how rubbish this question is. $X$ is not a set and $Mor(X,X)$ does not comprise of set mappings from $X$ to $X$. – user1992 Jun 10 '14 at 06:21
  • @user1992 See my answer, which I hope clarifies your doubts. –  Jun 13 '14 at 13:45

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$\mathbf{Mor}(X,X)$ is always a monoid. Rankeya defines $\mathbf{Mor}(X,X)$ to be $G$, a group. $G$ has inverses by definition; hence, each member of $\mathbf{Mor}(X,X)$ has an inverse.