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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

  1. The function f is injective if $f(x)=f(y)$ implies $x=y$ for all $x,y\in Dom(f)$
  2. The function f is surjective if for every $b\in Codom(f)$ there exists some $a\in A$ such that $f(a)=b$
  3. The function f is bijective if it is both injective and surjective



Exercise

Is $\Phi$ surjective, injective, bijective?


My Solution

  1. $\Phi$ 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 $\hspace{15cm}$ ${\Large ▫}$

  2. $\Phi$ 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

  1. Is the surjective proof correct?
  2. How can I define two functions in $\mathcal{F}$ such that $\Phi$ is not injective?

Can someone give me some hints/tips?

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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 $f(x_1)=y_0\ne y_1=g(x_1)$.

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  • $\begingroup$ Great, I get it now! Thank you $\endgroup$ – Onur Oct 22 '12 at 22:05

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