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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.

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en.wikipedia.org/wiki/… –  pedja Oct 13 '11 at 8:53
2  
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 gowers.wordpress.com/2011/10/11/… –  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

2 Answers 2

up vote 3 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 $(r_1,r_2)\in\mathbb{R}^2$.

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$g$ 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 $(a, b, 1)$.

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