2

I would like to know if every surjective group homomorphism $f:\mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ is also injective. I suspect it is true, but I'm not sure how to go about proving it.

If it's true, this would be an extremely useful fact for various homology calculations that appear in exercises throughout Hatcher's Algebraic Topology text!

Thanks!

michiganbiker898
  • 1,499
  • 2
    In addition to the link from the previous comment, Smith normal form will work more specifically for the case where you replace $\mathbb{Z}$ by a PID: $f$ is represented by a $2\times 2$ matrix over $\mathbb{Z}$, which can be written as $P D Q$ where $P, Q$ are unimodular matrices and $D$ is diagonal. $f$ is surjective resp. injective if and only if $D$ is; and $D$ being surjective implies the diagonal entries of $D$ are units, and therefore nonzero which implies $D$ is injective. – Daniel Schepler Dec 03 '21 at 05:08
  • 5
    To summarize the above link: $0\subsetneq$ker$(f)\subsetneq$ker$(f^2)\subsetneq$ker$(f^3)\cdots$ gives a strictly increasing infinite sequence of subgroups of $\mathbb{Z}\oplus\mathbb{Z}$, which is impossible. – tkf Dec 03 '21 at 05:10

0 Answers0