Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having a bit of trouble with injections and surjections. Injections are like one to one functions (for every element in the domain, there is one, and only one in the range) and surjections are functions that hit every part of the range I believe. How would one prove $g: \mathbb{R}^3 \to \mathbb{R}^2$ where say $g$ is defined as $$g(x,y,z) = (xz,yz)$$ is(n't) injective or surjective?

For injectivity, I believe providing the counter-example $x,y,z = 1$ and $x,y,z = -1$ is enough to disprove it. That would be because there would be two points on the domain that for the range are equal, namely $g(1,1,1) = (1,1)$ and $g(-1,-1,-1) = (1,1)$.

For surjectivity, I believe that it is true. I believe you can map out the entire range from $(-\infty , \infty)$. I don't know how to go about proving the statement though.

share|cite|improve this question… – pedja Oct 13 '11 at 8:53
You have got the definition of injectivity wrong. Injective means $f(x_1)=f(x_2)$ implies $x_1=x_2$, or in words: "for every $y$ in the range, there is at most one $x$ in the range that is mapped to $y$". I recommend reading Gower's excellent blog post about the topic… – Fredrik Meyer Oct 13 '11 at 9:34
Your statement of the meaning of injection is completely wrong, but your example is right. Because of the correct example, I am sure that you actually know the meaning of injection, just did not translate your geometric knowledge properly into words. – André Nicolas Oct 13 '11 at 11:09
up vote 4 down vote accepted

You can let $z=1$ and let $x$ and $y$ each range over $\mathbb{R}$. More explicitly, $g(r_1,r_2,1)=(r_1,r_2)$ for any $(r_1,r_2)\in\mathbb{R}^2$.

share|cite|improve this answer

$g$ is not injective since $$g(3, 15, 4) = (12, 60) = g(2, 10, 6),$$ that is two different points of $\mathbb{R}^3$ go the same point of $\mathbb{R}^2$ under the map $g$. As Danielle explained, $g$ is surjective since if you take any $(a, b) \in \mathbb{R}^2$, the exists an element $(x, y, z)$ of $\mathbb{R}^3$ such that $g(x, y, z) = (a, b)$, namely $(a, b, 1)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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