This is what I have so far
$f(a)=x$
$f(a)=y$
$f(a)=z$
$f(b)=x$
$f(b)=y$
$f(b)=z$
Am I doing this right? From this, I can conclude that no inverse exists.
This is what I have so far
$f(a)=x$
$f(a)=y$
$f(a)=z$
$f(b)=x$
$f(b)=y$
$f(b)=z$
Am I doing this right? From this, I can conclude that no inverse exists.
On
A single function must have an answer for every object in it's domain.
For example, if I had a domain of $\{a, b, c, d\}$ and a range of $\{x, y, z, w\}$ then a single function would be:
$$f_1(a)=x, f_1(b)=x, f_1(c)=x, f_1(d) = y$$
That's a single function that I just defined. I defined it by giving any value in the range for every single value in the domain.
A second function definition might be something like:
$$f_2(a)=w, f_2(b)=x, f_2(c)=y, f_2(d)=z$$
For a function to have an inverse means that if we were to use that first function definition I just gave and flip the domain and range, does the result behave like a function. Again using my first example.
$$g_1(x)=a, g_1(x)=b, g_1(x)=c, g_1(y)=d$$
But this doesn't make any sense because a function can't have multiple definitions for a single value. This is saying that $g(x)$ is equal to $a$ and $b$ and $c$ which doesn't work at all. This means that the function $f_1$ has no inverse.
We could go a step further and point out that there are no definitions for $g_1(z)$ or $g_1(w)$, however depending on the teacher/math philosophy, sometimes that last bit doesn't matter. You can get around this issue by doing some math hand waving and saying "well the image of $f$ was just $x$ and $y$ so the inverse only needs to have a domain of $x$ and $y$, not the full $\{x, y, z, w\}$."
I might check with your teacher, or check in your notes how "inverse" is defined and what exactly your teacher meant, because that would change your answer as to which functions have inverses.
You are going to have a total of $9$ functions. Each of $a$ or $b$ could be mapped to $x$,$y$,or$z$.
Out of these $9$ functions, $6$ of them are one-to-one.
Note that if $a$ is mapped to something, then $b$ can be mapped to the other two.
Therefore $a$ has three choices and for each of these choices $b$ has only two choices.
That makes a total of $3\times 2 =6$ one - to- one functions.