1
$\begingroup$

I had previously figured out injectivity/surjectivity on basic functions but I am stumpted when it comes to showing functions which are cartesian products are injective/surjective.

The first one: $$f: \Bbb{Z} \to \Bbb{Z}\times\Bbb{Z},$$ where $\Bbb{Z}$ is integers set.

$$f(n) = (2n, n+3)$$

I had shown that f was injective by contrapositive:

suppose $f(x) = f(y)$

then $(2x, x+3) = (2y, y+3)$

but I am unsure if this proof is complete..

When showing onto, I could not think of any counter examples. When dealing with basic functions I would just show that $f(x) = y$ but I am unsure how to show that with cartesian products.

$\endgroup$
  • $\begingroup$ Almost anything works to show the map is not surjective. For example, there is no $x$ such that $f(x)=(17,6)$ because $17$ is odd. There is no $x$ such that $f(x)=(0,0)$, since $f(x)=(2x,x+3)$ and if $2x=0$, then $x=0$. $\endgroup$ – André Nicolas Sep 13 '14 at 2:28
2
$\begingroup$

From $ (2x, x + 3) = (2y, y + 3) $ it follows that $ 2x = 2y $ and $ x + 3 = y + 3$, equalities which will only be true if $ x = y $.

Then you must notice (with a bit of given practice) the function is not surjective. The reason is every element in the range of $f$ is of the form $ (2n, n + 3) $ for some integer $n$. But there are plenty points on the lattice $ \Bbb Z \times \Bbb Z$ that are not of this form. Find one such point and assume there is $ x \in \Bbb Z $ for which $ f(x) = $ to the said point and try to arrive at a contradiction. One of the easiest one's to try is the point $( 0, 0 )$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.