Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to solve this exercise:

Consider $\psi: f\in \mathbb Z^{\mathbb Z} \mapsto f(2) \in \mathbb Z$. Is $\psi$ injective? Is it surjective?

I've never had to work with functions as elements before, so I am a bit confused. Does $f(2)$ mean a function $f$ having $2$ as parameter? Proving $\psi$ to be an injection means that I have to find a $g$ function so that $\psi(f)=\psi(g) \Leftrightarrow f=g$. How do I do that? And how do I prove that $\exists g \in \mathbb Z^{\mathbb Z} : \; \psi(g)=f(2)$? A nudge in the right direction would be really appreciated.

share|improve this question
1  
$f(2)$ is simply the value of $f$ at $2$. Example: if $f(n)=n^2-1$ then $f(2)=2^2-1=3$. –  Andrea Mori Dec 19 '12 at 10:17
add comment

2 Answers

up vote 2 down vote accepted

HINT: Answer these questions:

  • Let $f$ and $g$ be two functions on $\Bbb Z$. Suppose that $f(2)=g(2)$. Is it then true that $f=g$ as functions?

  • Let $n\in\Bbb Z$. Is there a function $f:\Bbb Z\rightarrow\Bbb Z$ such that $f(2)=n$?

share|improve this answer
    
On one hand $\psi$ is not an injection because $f(2x)=g(x^2) \text{ if } x = 2$, but that does not imply $f = g$. On the other hand $\exists f : \mathbb Z \rightarrow \mathbb Z \mid f(2)=n,\; n\in \mathbb Z$ and that is $f\left(\frac{1}{2}n\right)$, so $\psi$ is surjective. Am I correct? –  haunted85 Dec 19 '12 at 10:31
    
No, actually what you're saying doesn't make much sense. I believe that you have a confusion about the notation. To give a function $f$ with domain $\Bbb Z$ you have to specify the value that $f$ attains at each $n\in\Bbb Z$. The notation $f(n)$ simply denotes this value. See the example that I gave in the comment. –  Andrea Mori Dec 19 '12 at 10:36
    
Ok, I will try to rephrase. Let's consider $f(n)=n^2$ and $g(n)=2n$ then $f(2)=g(2)$ but $f \neq g$. So $\psi$ is not injective. Am I ok at least with this? Find hard to answer your second question. –  haunted85 Dec 19 '12 at 11:03
    
Ok, much better now! :) The second question is actually very trivial once you consider what a function really is. The difficulty is only psychological. –  Andrea Mori Dec 19 '12 at 11:06
1  
The point is that functions can be defined arbitrarily, in the sense that--in absence of other stringent conditions--you have no constraints in declaring that the value of a function at a given point is any value you want. See William's answer for a "simple" solution. –  Andrea Mori Dec 19 '12 at 12:01
show 2 more comments

$\Psi$ is not injective.

Let

$f(x) = \begin{cases} 0 & \quad x = 2 \\ 1 & \quad x \neq 2 \end{cases}$

and let $g$ be the constant $0$ function. Then $f \neq g$ but $\Psi(f) = f(2) = 0 = g(2) = \Psi(g)$.

$\Psi$ is surjective. Let $k \in \mathbb{Z}$. Let $g_k$ be the constant function taking value $k$. Then $\Psi(g_k) = k$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.