Given
Let $\mathcal{F}(X,Y)$ be the set of all functions $f:X\longrightarrow Y$, and let $x_0 \in X$. Given:
$$\Phi :\,\mathcal{F}(X,Y)\longrightarrow Y$$
Defined by $\Phi (f)=f(x_0)$ for $f\in\mathcal{F}(X,Y)$.
Definition
- The function f is injective if $f(x)=f(y)$ implies $x=y$ for
all $x,y\in Dom(f)$- The function f is surjective if for every $b\in Codom(f)$
there exists some $a\in A$ such that $f(a)=b$- The function f is bijective if it is both injective and
surjective
Exercise
Is $\Phi$ surjective, injective, bijective?
My Solution
-
$\Phi$ is surjective
Proof: let $y\in Y$. We will show that there exists some
element $x\in X$ such that $\Phi(f)=f(x)=b$. Define $f(x)=y, \forall
x\in X$, We know $f\in\mathcal{F}(X,Y)$, because $\mathcal{F}(X,Y)$
is the set of all functions from $X$ to $Y$ therefore
$\Phi(f)=f(x_0)=y$, thus $\Phi$ is Surjective
$\hspace{15cm}$ ${\Large ▫}$ -
$\Phi$ is not injective:
Proof: Special case If $|X|=1$, then
$\Phi$ is injective.But when $|X|>1$, $\Phi$ is not injective. Ok here I get stuck. I
cant figure out how to define two functions such that $\Phi$ isn't
injective.
My Question
- Is the surjective proof correct?
- How can I define two functions in $\mathcal{F}$ such that
$\Phi$ is not injective?
Can someone give me some hints/tips?
Best Answer
Your surjectivity proof is ok.
You observe that $\Phi$ is injective if $|X|=1$. Note that $\Phi$ is also injective if $Y=\emptyset$ or $|Y|=1$. (You should prove injectivity in these three cases)
If $x_1\in X$ and $y_0, y_1\in Y$ with $x_1\ne x_0$, $y_0\ne y_1$, you can define two functions $f,g\colon X\longrightarrow Y$, namely $f(x)=y_0$ and $$g(x)=\begin{cases}y_0&\text{if }x=x_0,\\y_1&\text{otherwise.}\end{cases}$$ Then $\Phi(f)=\Phi(g)=y_0$, but $f\ne g$ because $f(x_1)=y_0\ne y_1=g(x_1)$.