$\newcommand{\Z}{\mathbb{Z}}$
For my purposes, a ring is always assumed to be a ring with identity and so because of this my working definition of a ring homomorphism requires that $\phi(1_S) = 1_R$ if $\phi:S \rightarrow R$.
Problem: Show that there exists a unique ring homomorphism $\varphi:\Z \rightarrow R$ for any ring $R$.
I know that this is standard and has some answers here, but I'd like to know if my approach justifying this via free groups makes sense.
Attempt: Recall that $\Z \cong F_1$ where $F_1$ denotes the free group generated by a singleton. Note that for any $r \in R$ if we consider the set $S = \{1\}$ we have a set map $f:S \rightarrow R$ by $1 \mapsto r$ and by the universal property of the free group it follows that $f$ extends uniquely to a group homomorphism $\varphi:F(S) \rightarrow R$ for which $\varphi(s) = f(s)$. In the particular case where we let $f$ be the set map defined by $1 \mapsto 1_R$ it follows that there is a unique group homomorphism extended from $f$ satisfying $\varphi(1) = f(1) = 1_R$. Because any ring homomorphism must be a group homomorphism between the additive structures of $\Z$ and $R$ it follows from the uniqueness of this group homomorphism that any ring homomorphism must also be unique because by definition it must satisfy $\phi(1_\Z) = 1_R$.
How do I finish this argument off? I know I just need to show preservation of multiplicative structure under $\varphi$, and I know that because $\varphi:\Z \rightarrow R$ is a group homomorphism it follows that
$$
\varphi(n) = \varphi(\underbrace{1+\dots+1}_{n \text{ times}}) = \sum_1^n \varphi(1),
$$
which would also let us deduce that $\varphi(-n) = -\varphi(n)$. Is it just sufficient to say that $\varphi(ab) = ab(\varphi(1))$ which we can sequester into $a$ sums of $\varphi(1)$ and $b$ sums of $\varphi(1)$ or does this really require some type of case by case analysis depending on the signs of $a,b \in \Z$? Thanks in advance for the help and advice. Also of course let me know if my free group justification is wrong anywhere, it might be way too wordy because I'm not very comfortable with the material.
Best Answer
If you want to argue via $a = \sum_{i = 1}^a 1$ and $b = \sum_{i = 1}^b 1$, then you’ll have to make case distinctions because these formulas only hold for $a, b ≥ 0$.
We can take a different approach instead, which relies on the universal property of $ℤ ≅ F_1$: let us fix an arbitrary element $b$ of $ℤ$ and then consider the two auxiliary maps $$ h_b \colon ℤ \to R \,, \quad a \mapsto φ(ab) $$ and $$ k_b \colon ℤ \to R \,, \quad a \mapsto φ(a) φ(b) \,. $$ We need to show that $h_b = k_b$. Both maps are homomorphisms of abelian groups, i.e., additiv, because the multiplications in $ℤ$ and $R$ are distributive in the first argument and $φ$ is additive. It therefore suffices, by the universal property of $ℤ ≅ F_1$, to show that $h_b(1) = k_b(1)$. This equality holds because $$ h_b(1_ℤ) = φ(1_ℤ ⋅ b) = φ(b) = 1_R ⋅ φ(b) = φ(1_ℤ) ⋅ φ(b) = k_b(1_ℤ) \,. $$