MHB The Category Grp - Aluffi - Section 3.3 - basic question/clarification

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In Section 3.3 of Paolo Aluffi's "Algebra: Chapter 0," the function \( i_G: G \to G \) defined as \( i(g) := g^{-1} \) raises questions about notation and the role of identity elements versus inverses. The distinction between \( i_G \) and \( i \) is clarified, as both refer to the same function applicable to all groups without causing confusion. The discussion also highlights that any semi-group homomorphism between groups is inherently a group homomorphism, preserving both the identity element and inverses. This preservation is significant, as it contrasts with monoids, where semi-group homomorphisms do not necessarily maintain the identity element. The clarification emphasizes the unique properties of group homomorphisms in relation to their structure.
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I am reading Paolo Aluffi's book, Algebra: Chapter 0.

I am studying Chapter II: Groups, first encounter. In Section 3 of this chapter, Aluffi deals with the category Grp in which the objects are groups and the morphisms are group homomorphisms.

Section 3,3, which is a pause for reflection, reads as follows:View attachment 2668

I am somewhat confused by the specific function described in that Aluffi writes:

$$ i_G \: \ G \to G , \ \ i(g) := g^{-1} $$.I have two rather simple questions:

1. Why do we have $$ i_G $$ in one place and $$ i $$ in the other - that is, shouldn't the above read $$ i_G \: \ G \to G , \ \ i_G(g) := g^{-1} $$?

2. Aluffi mentions both the identity element and inverses, but the function he considers seems to only deal with inverses? What is going on?

I realize that these are pretty simple issues, but would appreciate someone clarifying the situation for me.

Peter

***EDIT*** I have been reflecting on the above and now feel, regarding question 2 above, that I misunderstood what Aluffi was saying - the function specified only referred to inverses.
 
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1. Yes. But the map $i$ and $m$ as well are defined the same way for all groups $G$, so no confusion should occur.

2. The point is: any SEMI-GROUP homomorphism between groups is also a group homomorphism, since for any homomorphism:

$\varphi: G \to G'$

we have:

$\varphi(e_G) = e_{G'}$

$\varphi(g^{-1}) = [\varphi(g)]^{-1}$

which high-lights something about "group-ness", just preserving the map $m$ (in the sense that:

$\varphi \circ m_G = m_{G'} \circ (\varphi \times \varphi)$)

ensures that $\varphi$ preserves $i$ and $e$ (which we can think of as a "special" map $e:1 \to G$):

$\varphi \circ i_G = i_{G'} \circ \varphi$

$\varphi \circ e_G = e_{G'}$

To see why this is special, note that a similar assertion is NOT TRUE for monoids: a semi-group homomorphism between monoids does NOT necessarily preserve the identity, and this condition must be stipulated as an additional condition to have a monoid homomorphism.
 
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