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.


1 Answer 1


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.

(Of course, now we need to make sure that this is actually defines a group homomorphism. It's the only thing that could give a group homomorphism, but that doesn't mean it must be a group homomorphism.)

  • $\begingroup$ Thanks, that makes sense. $\endgroup$
    – frelg
    Commented Jun 20 at 8:12

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