Homomorphism from $\mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}$ has non-trivial kernel: elementary argument One can give an elementary arguments (avoiding "rank") to prove that any group homomorphism $f$ from $\mathbb{Z}\oplus \mathbb{Z}$ to $\mathbb{Z}$ has non-trivial kernel: 
Let $f:(1,0)\mapsto a$ and $f:(0,1)\mapsto b$, and we can assume that $a,b\neq 0$. Then $(b,-a)$ is in the $\ker f$. 
I was wondering- can we give such elementary argument to prove that any homomorphism from $\mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}$ to $\mathbb{Z}\oplus \mathbb{Z}$ has non-trivial kernel? (I would like to avoid to use the "rank" etc.)

Likewise, in topology, $\mathbb{R}$ and $\mathbb{R}^2$ are not homeomorphic, and this can be proved by some "connectedness" argument. But, this argument may not work to prove that $\mathbb{R}^2$ and $\mathbb{R}^3$ are not homeomorphic. So, in above question, I don't know whether we can give elementary argument
 A: Suppose $(a,b),(c,d),(e,f)\in\Bbb Z^2$ satisfy no nontrivial relation. Then in particular, no pair of them does. Since $d(a,b)-b(c,d)=(ad-bc,0)$, if a pair of vectors in $\Bbb Z^2$ are independent then the determinant of the resulting square matrix must be nonzero. (Otherwise $b$ and $d$ would be $0$, which is impossible since $(a,0)$ and $(c,0)$ are not linearly independent.) We now have
$$\begin{array}{l} (ad-bc,0) = \color{Purple}{d(a,b)-b(c,d)}, \\ (af-be,0)=\color{DarkOrange}{f(a,b)-b(e,f)}. \end{array}$$
Now simplifying $(af-be)\big(\color{Purple}{d(a,b)-b(c,d)}\big)-(ad-bc)\big(\color{DarkOrange}{f(a,b)-b(e,f)}\big)$ gives
$$0=(cf-de)b\color{Red}{(a,b)}-(af-be)b\color{Blue}{(c,d)}+(ad-bc)b\color{Green}{(e,f)}.$$
Since the three determinants $cf-de$, $af-bc$, $ad-b$ are nonzero, for this to be a trivial relation we are forced to conclude $b=0$. By symmetry, we can interchange the roles of the three vectors in this argument and conclude similarly that $d=0$ and $f=0$. One can even swap the coordinates of each vector (which preserves linear independence) to get $a,c,e=0$ too. But this would imply the three vectors are $(0,0),(0,0),(0,0)$ which is impossible!
A: Suppose we have a map $f:\Bbb Z^n\to\Bbb Z^m$ that is injective. Since $\Bbb Q$ is $\Bbb Z$-flat, we obtain an injection of $\Bbb Q$-vector spaces $1\otimes f:\Bbb Q^n\to \Bbb Q^m$, so that $n\leqslant m$.
A: Yes, one can do a construction like the one you provided for $\mathbb Z\oplus\mathbb Z\to\mathbb Z$.
Let $f:\mathbb Z\oplus\mathbb Z\oplus\mathbb Z\to\mathbb Z\oplus\mathbb Z$ be any homomorphism, let $f(1,0,0)=(a_1,b_1)$, $f(0,1,0)=(a_2,b_2)$, and $f(0,0,1)=(a_3,b_3)$. Consider $$(a_2b_3-a_3b_2,-(a_1b_3-a_3b_1),a_1b_2-a_2b_1),$$
the cross product of the vectors $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$.
