Difficulty understanding this example that proves a function is one-to-one

Question

The function $f:\mathbb R\to\mathbb R$ given by $f(c) = 6x + 2$ is one-to-one because $$\begin{align} f(x_1) &= f(x_2) \\ \implies 6x_1 + 2 &= 6x_2 + 2 \\ \implies 6x_1 &= 6x_2 \\ \implies x_1 &= x_2. \end{align}$$

I understand that a one-to-one function is a function in which for each $y$ in the range of $f$, there is only one $x$ such that $f(x) = y$.

I understand that when you map out $f(x) = 6x + 2$, you get a linear equation when each value of $x$ corresponds to a unique $y$ value.

But this notation is really confusing me: $f(x_1) = f(x_2)$

If $f(x_1)$ and $f(x_2)$ are equal to each other then aren't they are pointing to the same $y$ value, so doesn't that contradict being one-to-one?

What exactly are they talking about when they refer to $x_1$ and $x_2$?

Answer 1

You understand that with a one-to-one function, for each $y$ there can be only one $x$ so that $f(x)=y.$ How do we check to see whether there really is only one $x$ so $f(x)=y$?

One way is to suppose $f(a)=y$ and $f(b)=y$ for some $a$ and $b$. If the function really is one-to-one, then $a$ must equal $b$. This is again, because only one value gets sent to $y$.

Your book is using $x_1$ and $x_2$ for variable names, instead of $a$ and $b$, but the idea is the same.

Answer 2

The author uses $x_1$ and $x_2$ to initially represent two distinct variables that are not necessarily related. Then he essentially says, "Let's say that $f(x_1)$ and $f(x_2)$ are equal and see what happens". From there he shows that these two random variables must be equal.

You could instead use $a$ and $b$ and it would serve the same purpose.

Answer 3

It's best to see this on a function which is not one to one. Imagine that we had a function $f(x)=x^2$. Then it's the case that $f(3)=f(-3)$. Now we have two different points $(3,f(3))$ and $(-3,f(-3))$ which have the same $y$-value.

For a one-to-one function. We want to show that this type of thing does not happen. In fact we want a criteria to make sure this doesn't happen. So we say that whenever two $y$ values are the same: $f(x_1)=f(x_2)$ that these are infact not two different points on our function: because there x values are the same. And the we see that $(x_1,f(x_1))$ is the exact same point on the graph as $(x_2,f(x_2))$.

Answer 4

one-one(injective) function maps distinct points of domain to distinct points of codomain. So

If $f:X→Y$ is one-one then, for all $x,y$ in $X$ we must have,

$$\begin{align} f(x)≠f(y)\implies x≠y\\ \iff f(x)=f(y)\implies x=y \end{align}$$