homomorphisms on $\mathbb Z\oplus \mathbb Z$ Is it true that a group homomorphism $f:\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ that is injective must be an isomorphism?
I know that non zero homomorphisms $g:\mathbb Z\to \mathbb Z$ are all injective but are isomorphisms only if $g(1)=\pm 1$. But the situation for homomorphisms of
$f:\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ is not clear to me.. I know that $f $ is completely determined by $f(1,0)$ and  $f(0,1)$ 
 A: I believe $f(a,b)=(2a,2b)$ is an easy counterexample.
It is easy to see that this map is injective group homomorphism, but it is not surjective.

As Arturo Magidin pointed out in his comment, for any injective homomorphism $f:\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ the image $\operatorname{Im} f$ is isomorphic to $\mathbb Z\oplus \mathbb Z$, i.e., it is a free Abelian group of rank 2. Even more, for any homomorphism $f:\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ the image is generated by $f(1,0)$ and $f(0,1)$, which implies that it will be a free Abelian group. (It is explained more detailed in the comments.)
A: From the comments, it seems that the actual situation is:

Suppose that $f,h\colon\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$ are group homomorphisms, and that $h$ is a retraction of $f$; that is, $h\circ f = \mathrm{id}$. Is $f$ an isomorphism?

The answer in that situation is "yes". 
Since $h\circ f$ is bijective, $f$ is one-to-one and $h$ is onto. It is also straightforward to verify that 
$$\mathbb{Z}\oplus\mathbb{Z}\cong \mathrm{Im}(f) \oplus \mathrm{ker}(h).$$
Both $\mathrm{Im}(f)$ and $\mathrm{ker}(h)$ are free abelian. Since $f$ is one-to-one, $\mathrm{Im}(f)$ is free abelian of rank $2$. Since the rank of a free abelian group is uniquely determined, this means that
$$2 = \mathrm{rank}(\mathbb{Z}\oplus\mathbb{Z}) = \mathrm{rank}(\mathrm{Im}(f))+\mathrm{rank}(\mathrm{ker}(h)) = 2 + \mathrm{ker}(h),$$
so $\mathrm{ker}(h)$ is trivial. Therefore, $h$ is one-to-one and onto, hence bijective, and since $h\circ f = \mathrm{id}$, it follows that $f$ is the inverse of $h$, and an isomorphism. 
A: A more interesting exercise is what condition on $f(1,0) = (a,b)$ and $f(0,1) = (c,d)$  is needed for the homomorphism $f: {\mathbb Z} \oplus {\mathbb Z} \to {\mathbb Z} \oplus {\mathbb Z}$ to be an isomorphism?  The answer is that $ad - bc = \pm 1$ is necessary and sufficient.  Now generalize to homomorphisms of ${\mathbb Z}^n$ to itself.
A: To follow up on Robert Israel's answer, consider the $n$-fold direct sum $R^{n}$ where $R$ is a commutative ring with $1$. A homomorphism $f:R^{n} \rightarrow R^{n}$ corresponds to an $n \times n$ matrix $M$ with entries in $R$. It turns out that $f$ is injective if and only if $\det (M)$ is a regular element (i.e., non zero divisor) in $R$ and an isomorphism if and only if $\det (M)$ is a unit of $R$, whence Robert's condition. So, every injective $f$ is an isomorphism if (and only if) every regular element of $R$ is a unit. Such are rings are often called "quorings." 
