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I require some clarification and hints on the following Problem: \begin{align}f: X \longrightarrow Y \end{align} The image $p$ is defined as: \begin{align} p: G_f &\longrightarrow X \\ (x,y) &\longmapsto x\end{align} With $G_f=\lbrace (x,y) \in X \times Y : f(x)=y \rbrace$

Question: Is the function $p$ a bijection?

The part I struggle the most with is that this is the first time that I deal with a 2-tuple, so please check if my approach is correct.

In the general case, to show if a function is injective choose $x,x' \in X$ such that $f(x)=f(x')$ and show that if $f(x)=f(x')$ then $x=x'$

Here is my approach to this problem: Let $(x,y) \in G_f \subset X \times Y$ and $(x',y') \in G_f \subset X \times Y$ be two 2-tuples such that $p(x,y)=x=p(x',y')$ then $(x,y)=(x',y')$ is an ordered pair, therefore $x=x'$ and $y=y'$

Is this correct?

For surjectivity, no ideas so far.

Surjectivity: $ \forall y \in Y \exists x \in X : f(x)=y $

Spaced
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  • Are you asking if $p$ is a bijection? Or if $f$ is a bijection? It doesn't seem like you've put any conditions really on $f$. – user2566092 Sep 27 '13 at 20:05
  • yes, I will edit this into the header. I am asking if $p$ is a bijection. I did copy the problem from my book, where they kept the statement about $f$. I thought it might have an impact on $G_f$. – Spaced Sep 27 '13 at 20:07

1 Answers1

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Assume $p(x,y)=p(x',y')$. Then $x=x'$ and hence $(x',y')=(x,y')$.

This shows that $(x,y)$ and $(x,y')$ are points on $G_f$ and by definition

$$y=f(x) , y'=f(x) \,.$$

This shows that $y=y'$.

P.S. In your proof, the part $p(x,y)=x=p(x'y')$ then $(x,y)=(x'y')$ is incomplete, that's exactly what you try to prove. $p(x,y)=x=p(x'y')$ only implies that $x=x'$, you need to also argue that $y=y'$, which is not hard.

N. S.
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  • Thanks for your response, I tried to argument through the statement of ordered pairs, we once had to prove that $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$ I thought this might would work here as well. – Spaced Sep 27 '13 at 20:21
  • Is there any specific reason behind the argument that if $(x,y)=(x',y')$ only implies that $x=x'$ but not $y=y'$? – Spaced Sep 27 '13 at 20:32
  • @Spaced That's not what I said. The issue is : Why do you get that $(x,y)=(x',y')$? Note that all you know is that $p(x,y)=x=p(x',y')$, how do you conclude from here that $(x,y)=(x',y')$? – N. S. Sep 27 '13 at 20:35
  • I see, I get this now, thanks for your assistance. – Spaced Sep 28 '13 at 11:10