Is every well-defined map continuous?  I am working on the function $f:[0,1) \to [0,1]$ defined by $f(x)=\dfrac{x}{x+1}$ for each $x\in [0,1)$.
It is clear that $f$ is well defined and continuous.
Now consider another function given by
$$
F(x)=
 \begin{cases}
 f(x), &x\in[0,1),
 \\ 1, & x = 1.
 \end{cases}
$$
It is clear that $F$ is an extension of $f$ into $[0,1]$. But it is not clear to me that $F$ is not continuous at $x=1$, even though it is defined at $x=1$.
Please I need more clarification about this. 

In general, a function $g$ is well-defined at $a$ if $g(a)$ exists.
In general, a function $g$ is continuous at $a$ if:


*

*$g(a)$ exists, and

*$\lim\limits_{x\to a} g(x)$ exists, and

*$g(a) = \lim\limits_{x\to a}g(x)$.


Therefore, if a function $g$ is continuous at $a$, then it must be well-defined there.  However, the converse may not be true: a function may be well-defined at a point, but not continuous there.
For example, your function $F$ is well-defined at $x = 1$ (since $F(1)$ exists), but is not continuous at $x = 1$ (since $\lim\limits_{x \to 1}F(x) \neq F(1)$.)
On the other hand, note that while your function $f$ is well-defined and continuous on $[0,1)$, it is not well-defined at $x = 1$, and so it is not continuous at $x = 1$.
