What are the other methods used to prove that a homomorphism is bijective?

Question

The motivation can be found in: Show that $ℤ^{m}$ is a subgroup (and a free abelian group) of $ℤ^{n}$ for all $m≤n$.

In a specified problem related to a dynamical system the only possibility is $m=n$. Here $m$ and $n$ have special meanings: $m$ is the cardinal of a population $A$ and $n$ is the cardinal of a population $B$.

I want to construct a bijction between $\mathbb{Z}^m$ and $\mathbb{Z}^n$ (I mean that I try to prove that $m=n$). I have proved that $m≤n$. So there is an injective homomorphism between $\mathbb{Z}^m$ and $\mathbb{Z}^n$. I want now to prove that this map is also surjective. However, I have no idea to start.

My question is: What are the other methods used to prove that a homomorphism is bijective? I have tried to use a non-degenerate quadratic form. But I have no idea to start.

Also, I'll need the identity elements of $\mathbb{Z}^m$ to be mapped to the identity of $\mathbb{Z}^n$. But, there is some related problems in this way.

Answer 1

Here's a thought. Suppose that you have an injective homomorphism $f : \mathbb{Z}^m \to \mathbb{Z}^n $ and you want to prove that $m=n$. While it would be sufficient to prove that the homomorphism is surjective, that is not necessary. Here are two necessary and sufficient conditions for $m=n$, which are useful in different situations:

• $image(f)$ has finite index in $\mathbb{Z}^n$
• $image(f)$ is "cobounded" in $\mathbb{Z}^n$, meaning that there exists a constant $D \ge 0$ such that for each $p \in \mathbb{Z}^n$ there exists $q \in \mathbb{Z}^m$ such that $d(p,f(q)) \le D$.

Answer 2

I have used the Lee Mosher conditions to poste this solution. However, there a litle problem with it.

If $m≤n$, then we can embed $ℤ^{m}$ inside $ℤ^{n}$ in the obvious way by the map $$θ₁:ℤ^{m}→ℤⁿ:θ₁(z₁,…,z_{m})=(z₁,…,z_{m},0,…,0)$$

I will use the second equivalent form in the Lee Mosher answer.

There exists a constant $D≥0$ such that for each $p=(p₁,....,p_{n})∈ℤⁿ$ there exists $q=(q₁,....,q_{m})∈ℤ^{m}$ such that $$d((p₁,....,p_{n}),(q₁,....,q_{m},0,...,0))≤D$$

We have $$d((p₁,....,p_{n}),(q₁,....,q_{m},0,...,0))=√(∑_{i=1}^{n}(p_{i}-q_{i})²)=√(∑_{i=1}^{m}(p_{i}-q_{i})²+∑_{i=m+1}^{n}p_{i}²)$$

We can choose $$(q₁,…,q_{m})=(p₁,…,p_{m})$$

So we get $$d((p₁,....,p_{n}),(q₁,....,q_{m},0,...,0))=√(∑_{i=m+1}^{n}p_{i}²)$$

This quantity does not exist since we want to prove that $m=n$.