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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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    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.2011-10-13

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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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Yes, the definition of an injection is wrong. The definition provided seems to be the traditional definition of a standard singular function.

These generally, functions, or real valued maps, are not many valued (which is generally what is meant by function) always satisfy the requirement that every element in the domain is mapped to one and only one element of the range, but not the conversely. they need to preserve distinct-ness or be injective.every domain element attains a distinct element of the co-domain.,

. In-jectivity relates to the fact, that no two elements of the domain ever-correspond to the same element of the co-domain. It preserves distinct-ness

ie (injective-function) $F(x)=F(y) \leftrightarrow x=y$ and every element of the co-domain is corresponds to at most one element of the domain; distinct elements of the domain have a distinct images.

injective-functions are such that every element of the domain corresponds to one and only one element of the co-domain,(which is a property that function have in general) and in addition, in-jective function have this addition property that no function value can be repeated distinct domain values correspond to distinct function values, the function only takes the same function value if it is the domain point, if $x\neq y\to $F(x)\neq F(y);

The restriction of the injective function F$ from its domain to its image:

$F1:\text{dom}(F)\to \text{Im(F)}$

$IM(F)\subseteq \text{co-dom}(F)$ where $y\in\text{Im(F),are those}\, y\in \text{codom}(F)$ ,for which there $\exists x \in \text{dom}(F): F(x)=y

The inverse of this restricted function is defined and as in-jective uniquely so,

That this the restriction that part of the co-domain which correspond to an element of F's domain, they are defined.

which are is be sur-jective by definition, for a function, and thus F1 will be bi-jective , as its also in-jective

But the original function F$ $F:\text{dom}(F)\to\text{co-dom(F)}$

may not be not be surjective, as it may not have defined inverese for every element of the $\text{codom(F)}. some of those points may not correspond to an element of the domain.

, noR will this restriction F1$ be necessarily an interval; otherwise strictly mon-0tone functions which have bi-jective strictly increasing restrictions, would count as continuous,as image would be interval, which is not always so. Their inverse function may have an ill defined domain (not necessarily a closed interval)

whilst a Function generally only has this horn $ x=y \rightarrow F(x)=F(y)$ (and for every element of the domain it is mapped to at one, and only one element of the co-domain) to