# Unique group map out of free abelian group on n elements

For a set $$A$$, the tuple $$(F(A),j:A\to F(A))$$ (where, $$F(A)$$ is an abelian group and $$j:A\to F(A)$$ is a set map) has the universal property of a free abelian group on $$A$$ if for every abelian group $$G$$ and every set map $$f:A\to G$$, there is a unique group map $$\varphi:F(A)\to G$$ st $$\varphi\circ j=f$$.

Aluffi in Chapter 0 (Claim 5.4) uses the above stated universal property of free abelian groups to show that the the free abelian group on $$A=\{1,....,n\}$$ is $$(\mathbb Z^{\oplus n},j:i\mapsto e_i)$$ where $$e_i\in \mathbb Z^{\oplus n}$$ is $$0$$ everywhere except at the $$i$$-th position.

For any $$(G,f:A\to G)$$ as in the setup of the universal property, we want to show that there is a unique group map $$\varphi:\mathbb Z^{\oplus n}\to G$$. Since we want $$\varphi\circ j=f$$, we require $$\varphi(j(i))=\varphi(e_i)=f(i)$$. Until this part, everything in the proof is clear.

But what I don't fully get is why this is extended to all of $$\mathbb Z^{\oplus n}$$ as follows: $$\varphi\left(\sum\limits_{i=1}^nm_ij(i)\right)=m_i\sum\limits_{i=1}^{n}f(i).$$

When I did the proof by myself, I too chose to extend $$\varphi$$ in this way, but only later did it start to bother me as to why this is the “right” way to do it? I recognize that $$\mathbb Z^{\oplus n}$$ and $$G$$ are abelian groups and hence $$\mathbb Z$$-modules and $$\varphi$$ (as defined/extended above) is a $$\mathbb Z$$-linear map (also that such a map is indeed particularly a group map).

Can someone please help me see what is happening here? It might have to do with the structure preserving maps between abelian groups possibly being $$\mathbb Z$$-linear maps rather than just group homomorphisms.

You want $$\varphi$$ to be a group homomorphism. That means that if you know what you want $$\varphi(x)$$ and $$\varphi(y)$$ to be, then you are forced to make $$\varphi(x+y)$$ equal to $$\varphi(x)+\varphi(y)$$. That is the only way in which $$\varphi$$ has any hope of being a group homomorphism. Likewise, since $$mx$$, $$m\in\mathbb{Z}$$, $$m\gt 0$$, is just shorthand for $$\underbrace{x+\cdots+x}_{m\text{ summands}},$$ it follows that for $$\varphi$$ to have any hope of being a group homomorphism, we must have $$\varphi(mx) = \varphi\left(\underbrace{x+\cdots+x}_{m\text{ summands}}\right) = \underbrace{\varphi(x)+\cdots+\varphi(x)}_{m\text{ summands}} = m\varphi(x).$$ And for $$m\lt 0$$, we know $$mx = -\bigl( (-m)x\bigr)$$, then \begin{align*} \varphi(mx) &= \varphi\bigl( -(-m)x\bigr) = -\varphi((-m)x)\\ &= -\bigl( (-m)\varphi(x)\bigr) =(-(-m))\varphi(x)\\ &= m\varphi(x). \end{align*}
Since we have decided what we want $$\varphi(e_i)$$ to be for each $$i$$, there really is no choice about what $$\varphi$$ needs to be at $$\sum_{i=1}^n m_ie_i$$ for $$\varphi$$ to have any chance of being a group homomorphism. It has to satisfy $$\varphi\left(\sum_{i=1}^n m_ie_i\right) = \sum_{i=1}^n \varphi(m_ie_i) = \sum_{i=1}^n m_i\varphi(e_i).$$ That is the only definition that has any chance of being a group homomorphism.