Definition of surjective - understanding notation In Measures, Integrals and Martingales by René L. Schilling surjective (or onto) is defined as:
$$f(X) := \{f(x) \in Y\,:\,x\in X\} = Y$$
I think I understand the concept of surjective in a function $X \rightarrow Y$, but I am stuck with the notation. How can I read from left to right the math line above? In particular I am inquiring about:


*

*$:=$ Is it "defined as..."? If so, what are we defining?

*$:$ Is the colon read as "such that"?

*$=$ Is $X = Y$? It wouldn' make sense, would it? The first is in the domain, while the second in the range of the function.

*The use of upper and lower case. In this regard, notice how he defines an injective mapping:


$$f(x) = f(x')\Rightarrow x= x'$$
 A: The concept is defined in two steps.
First we define the image of a function $f : X \to Y$. The image of $f$, noted $f(X)$ is defined by $$f(X) := \{f(x) : x \in X\}.$$ Note that many people don't like the notation $:=$ and simply use the equality.
Now we can define the concept of surjection. A function $f : X \to Y$ is surjection if $$f(X) = Y.$$ In other words, for each $y \in Y$ you can find $x \in X$ such that $f(x) =y$.
A: The image $f(X)$ of $f$ is defined to be the set of points $f(x)$ in $Y$, where $x$ ranges over all of $X$. The function $f$ is surjective if this image $f(X)$ is all of $Y$, which is to say each $y$ in $Y$ is given as $f(x)$ for some $x$ in $X$.
added: the middle colon in the set means "such that." I have added some set braces to make your question clearer; because braces group things in TeX, you need a backslash before each to make them display. You are correct that "$:=$" is used to write "is defined as."
A: Here $f$ is assumed to be a function from $X$ to $Y$.  $X$ is the domain, $Y$ is the codomain, which does not have to be the same as $X$.
This statement can be unpacked as follows: 


*

*$f(X)$ is defined to be the set of $f(x)$ (which is a member of $Y$) for all $x$ in $X$.

*$f$ is surjective if $f(X) = Y$, i.e. every member of $Y$ must be $f(x)$ for some $x$ in $X$.


BTW,  René Schilling is a he, not a she.
