What is the equality of two functions (or maps)? Suppose there are two maps $f : A \to B$ and $g : A\to C$ such that $C$ and $B$ can be different sets, but $f(x)=g(x)$ for every $x$ in $A$. Thus, the image of $f$ and $g$ are both contained in the intersection of $B$ and $C$. 
Then, can it be said that $f=g$?  This seems like a trivial question, but really annoys and confuses me... 

Here, the function $F$ is defined as $F : M \to N$ but also regarded as $F : M \to F(M)$ to be a homeomorphism. So it still leaves confusion... 
 A: 
$f: \mathbb{N} \longrightarrow \mathbb{N}$ and $g: \mathbb{N} \longrightarrow \mathbb{N}\cup\lbrace \frac{1}{2} \rbrace$ defined by $f(x) = g(x) = x$ for all $x \in \mathbb{N}$. Note that $f$ is bijective but $g$ isn't.

$f:A \longrightarrow$ B and $g: C \longrightarrow D$ are equals when $A = C$, $B = D$ and $f(x) = g(x)$ for all $x \in A$ (by definition), otherwise, two equal functions may not have the same properties.
When you write $f:X \longrightarrow Y$ and $f: X \longrightarrow f(X)$ you are only considering the image, because the other elements of $Y$ are expendable. Writting in function of the image sometimes is convenient. In fact, they are two different things, for example:

$f: X \longrightarrow Y$ where $Y \subset \mathbb{N}$ and $f$ is injective. To show that $X$ is countable, we use that $f: X \longrightarrow f(X)$ is bijective and $f(X) \subset Y \subset \mathbb{N}$ is countable.

This is a abuse of notation, because $f: X \longrightarrow Y$ and $f: X \longrightarrow f(X)$ are two different things, but we keep the same name (makes the text more clear) because we only discard the elements that don't have any association.
A: They are not equal. For two maps to be equal you also require them to have the same domain and codomain.
What you can do for example, is "corestrict" f and g to $B \cap C$.
So it is true, that $f|^{B\cap C} = g|^{B \cap C}$.
A: In order for two functions to be equal, their images being the same is not enough, their co-domains also have to be the same.
For example, let $f: \mathbb{Z} \to \mathbb{Z}$ and $g: \mathbb{Z} \to \mathbb{C}$ with $f(x) = x$ and $g(x) = x$. Then notice that $f$ is a bijection while $g$ is not even surjection, hence not a bijection. So even some of the characteristics of functions change when we change the co-domain so we can't say $f= g$.
A: A function has three parts.
Domain, Codomain, and the assignment of $f:A\to B $.
Thus $f:A\to B$ and $g:C\to D$ are equal iff $ A=C$ and $B=D$ and for every $x\in A,$ $ f(x)=g(x).$ 
