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** 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.

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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
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 $\boxdot$

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