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?