Answer
a. **Arrow diagram**:
\(a \rightarrow v\)
\(b \rightarrow v\)
\(c \rightarrow t\)
b.
**\(f(A) = \{v\}\).**
**\(f(X) = \{v, t\}\).**
**\(f^{-1}(C) = \{c\}\).**
**\(f^{-1}(D) = \{a,b\}\).**
**\(f^{-1}(E) = \varnothing\).**
**\(f^{-1}(Y) = \{a,b,c\}\).**
Work Step by Step
## 1. The Sets and the Function
- **Domain**: \(X = \{a, b, c\}\)
- **Codomain**: \(Y = \{r, s, t, u, v, w\}\)
- **Function** \(f: X \to Y\) is given by:
\[
f(a) = v, \quad f(b) = v, \quad f(c) = t.
\]
### (a) Arrow Diagram
An arrow diagram lists the elements of \(X\) on the left and the elements of \(Y\) on the right, with arrows showing how each element of \(X\) maps into \(Y\). Here:
\(a \longrightarrow v\)
\(b \longrightarrow v\)
\(c \longrightarrow t\)
No other arrows come out of \(a,b,c\), since each has a single image under \(f\).
---
### (b) Finding \(f(A), f(X)\) and the Various Preimages
We have the following subsets:
\(A = \{a,b\}\)
\(C = \{t\}\)
\(D = \{u,v\}\)
\(E = \{r,s\}\)
\(Y = \{r,s,t,u,v,w\}\) (the entire codomain)
Recall the definitions:
\(f(A) = \{\,f(x) : x \in A\}\).
\(f^{-1}(S') = \{\,x \in X : f(x) \in S'\}\) for any \(S' \subseteq Y\).
Now we compute each required set.
#### 1. \(f(A)\)
\(A = \{a,b\}\).
\(f(a) = v\) and \(f(b) = v\).
Hence \(f(A) = \{v\}.\)
#### 2. \(f(X)\)
\(X = \{a,b,c\}\).
\(f(a) = v, \quad f(b) = v, \quad f(c) = t.\)
Hence \(f(X) = \{v, t\}.\)
#### 3. \(f^{-1}(C)\)
\(C = \{t\}\).
We want all \(x \in X\) such that \(f(x) \in \{t\}\).
Since \(f(c) = t\), the only element mapping into \(t\) is \(c\).
Thus \(f^{-1}(C) = \{c\}.\)
#### 4. \(f^{-1}(D)\)
\(D = \{u,v\}\).
We want all \(x \in X\) such that \(f(x)\) is either \(u\) or \(v\).
We have \(f(a) = v\), \(f(b) = v\), and \(f(c) = t\).
Only \(a\) and \(b\) map to \(v\) (which is in \(D\)); \(c\) does not.
Hence \(f^{-1}(D) = \{a,b\}.\)
#### 5. \(f^{-1}(E)\)
\(E = \{r,s\}\).
We want all \(x \in X\) such that \(f(x) \in \{r,s\}\).
Since \(f(a) = v\), \(f(b) = v\), and \(f(c) = t\), none of them is \(r\) or \(s\).
Therefore \(f^{-1}(E) = \varnothing.\)
#### 6. \(f^{-1}(Y)\)
\(Y = \{r,s,t,u,v,w\}\) is the entire codomain.
By definition, \(f^{-1}(Y)\) is all \(x \in X\) whose image lies in \(Y\). But **every** image of an \(x \in X\) must lie in \(Y\), because \(Y\) is the codomain.
Therefore \(f^{-1}(Y) = X = \{a,b,c\}.\)
