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.

** proof under construction - will post when done and more or less confident it's true.

** also please easy with the downgrades.. i don't understand why i'm getting them.


what is meant by show that?

am i supposed to give an example? sure g(y) can be y/2 if f is x*2. am i supposed to give a proof ? (we are learning the axioms and the lemmas.) in which case sure, again, given the sets A, B, X, Y and g: Y -> X, and f an injective function defined f: A->B with A a subset of X and B a subset of Y. we know that f will map only specific values of X to specific values of Y, i then define f(x) = x*2 and g(y) = y/2 thus g o f = idX is valid.

(i am not sure if this counts as a correct proof, but i am trying)

I can explain it verbally. I understand the concept. but i have NO idea what the question wants from me. "show that" is too vague.

share|improve this question
2  
It means "Prove that a function $f: X \to Y$ is injective if and only if there exists a $g: Y \to X$ such that $g \circ f = \mbox{id}_X$". Usually "Show that" means "Prove that". –  wj32 Nov 8 '12 at 5:24
    
oh thank you. i needed confirmation i can now proceed with a clear conscience. –  Shokodemon Nov 8 '12 at 5:34
    
When you finish your proof, you should post it here as an answer and accept it (if it's correct). –  wj32 Nov 8 '12 at 5:34
    
definitely will. –  Shokodemon Nov 8 '12 at 5:54
1  
The statement is wrong if $X$ is empty and $Y$ nonempty. –  Michael Greinecker Nov 18 '12 at 11:35

1 Answer 1

up vote 1 down vote accepted

Premiss: For $f: X \rightarrow Y$ $\wedge$ X,Y non-empty
Proposition: $f$ is injective $\iff$ there is a $f: Y \rightarrow X$ with $g \circ f:$$id_{x}$
Proof: {a direct proof}

In two parts; we show

a) $\Leftarrow$: for $f(x_{1}) = f(x_{2})$, f is injective,
b) $\Rightarrow$: it follows then with a) that an $f: Y \rightarrow X$ with $g \circ f:$$id_{x}$ exists.

  1. Then for a):

    Assume $f(x_{1}) = f(x_{2})$.
    Then $x_{1}$ $=$ $id_{x_{1}}$ $=$ $g \circ f(x_{1})$ $=$ $g \circ (x_{2})$ $= id_{x_{2}} = x_{2}$.

  2. for b):

    Assume $g: Y \rightarrow X$
    Then by the definition of a relation: $y_{1}\in$$f(X)$ $\wedge$ $y_{2}$$\in$[Y$\diagdown$$f(X)]$ And by definition of $g$ alongside $f$'s injectivity we have $g(y_{1}) = f^{-1}(y_{1}) = x_{1}$ $\Rightarrow$ $g \circ f=$$id_{x}$

QED $\boxdot$

share|improve this answer

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.