Tangent and arctangent functions: one-to-one? Is it correct to state that arctangent function is one-to-one, but the tangent function is not? Or can this only be stated with imposed restrictions?
 A: Injectivity and surjectivity are not merely properties of a mapping:  they are also properties of the domain and codomain on which the mapping acts.
In other words, if I say, let $f(x) = x^2$, is this a one-to-one mapping?  Recall that a function is one-to-one (injective) if for any $x \ne y$, $f(x) \ne f(y)$.  Well, I can't answer that question without further information because the domain is not specified.  If I say this is a function from the positive integers to the positive integers; i.e., $f : {\mathbb Z}^+ \to {\mathbb Z}^+$, then yes, it is injective.  But if I say this is a function from the reals to the reals, $f : {\mathbb R} \to {\mathbb R}$, then no, because $f(-x) = f(x)$ for $x \ne 0$.
There is a characterization of injective (one-to-one), surjective (onto), and bijective mappings that emphasizes the role of the domain and codomain:
A mapping $f : A \to B$ is injective if, for every $b \in B$, there is at most one $a \in A$ such that $f(a) = b$.
A mapping $f : A \to B$ is surjective if, for every $b \in B$, there is at least one $a \in A$ such that $f(a) = b$.
A mapping $f : A \to B$ is bijective if, for every $b \in B$, there is exactly one $a \in A$ such that $f(a) = b$.
So, if we specifically talk about the tangent and its inverse function, the issue here is how we specify the function and its domain and codomain.  For instance, if we say $$f : \mathbb R \to \mathbb R, \quad f(x) = \tan x,$$ then $f$ is not injective since $\tan x = \tan(x + \pi)$.  If we say $$f : {\textstyle(-\frac{\pi}{2}, \frac{\pi}{2})} \to \mathbb R, \quad f(x) = \tan x,$$ then yes, $f$ is injective.  Conversely, $$g : \mathbb R \to \mathbb R, \quad g(x) = \tan^{-1} x$$ is not surjective if we wish $g$ to be single-valued:  any choice of branch of the inverse tangent function will mean that not every real number in the codomain will have a preimage.  A more intuitive example of this is the previous function we discussed $f : \mathbb R \to \mathbb R$, $f(x) = x^2$.  This is not surjective since there is no real $x$ such that $f(x) = -1$.  To make it surjective, we can either expand the domain to $\mathbb C$, or restrict the codomain to $\mathbb R^+ \cup \{0\}$.
A: Let $\infty$ be neither $+\infty$ nor $-\infty$ but rather the $\infty$ that is approached by going in either the positive or the negative direction.  Or, the the language of topology, let $\mathbb R \cup\{\infty\}$ be the one-point compactification of $\mathbb R$.
Then the tangent function can be viewed as a one-to-one continuous function from $\mathbb R/2\pi$ onto (onto, not just into) $\mathbb R\cup\{\infty\}$. It's a homeomorphism from one (topological) circle to another.
But the tangent function can also be viewed as a periodic function from $\mathbb R$ onto $\mathbb R\cup\{\infty\}$, and then of course it is not one-to-one.
