How to determine whether the function is one-to-one? Differential Calculus Determine whether the function f is one-to-one 
f(t) is the number of people in line at a movie theater at time t. 
 A: Informally, we can think of a one-to-one function as one that maps distinct elements in the domain to distinct elements in the codomain.  Or, in other words, if $f$ maps $a$ and $b$ map to the same thing, then $a=b$.
Formally, a function $f:A \rightarrow B$ is called one-to-one if $f(a)=f(b)$ implies $a=b$.  Equivalently, if $a \neq b$, then $f(a) \neq f(b)$.
In this question, we have a function $f:T \rightarrow \mathbb{Z}^{\geq 0}$ defined by $f(t)$ is the number of people in line at a movie theater at time $t$, and $T$ is the set of times for which "time" is defined.  The task is to find two distinct times $t_1 \in \mathbb{R}$ and $t_2 \in \mathbb{R}$ for which $f(t_1)=f(t_2)$.
Here's a simple mathematical answer; it assumes (a) the theater has been open longer than an instant, and (b) there are a finite number of people in existence.


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*Suppose there are $N<\infty$ people in existence.  We pick $N+1$ distinct points of time $t_1,t_2,\ldots,t_{N+1}$ (assuming the theater has been open longer than an instant, since time is continuous, such points of time exist).  Then the pigeonhole principle implies there are two points in time $t_i$ and $t_j$ in which $f(t_i)=f(t_j)$; in other words, we have $N+1$ numbers $$f(t_1),f(t_2),\ldots,f(t_{N+1})$$ that all belong to $\{1,2,\ldots,N\},$ so they can't all be distinct.

