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Let $f\colon X \to Y$ and $g\colon Y \to X$ be functions. Assume $g \circ f$ is bijective. Prove $f$ is injective and $g$ is surjective.

Approach:

if $g \circ f$ is bijective then $g \circ f$ is one to one
if $g \circ f$ is bijective then $g \circ f$ is onto

so we know $$g ◦ f (a_1)=g ◦ f (a_2) \text{ this implies $a_1=a_2$}$$ $$\forall b\in X, \exists a\in x \text{ such that } g ◦ f(a)=b$$

so we have $$g(f(a_1))=g(f(a_2))$$ $$g(f(a))=b$$

From here, I am stuck. What's the next thing to consider?

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  • $\begingroup$ Please, consider accepting one of the anwers if your question has been addressed. (If not, it may lead to automatic deletion of your question after some time; but more to the point, this is the normal way of signaling your question has found an answer and is no longer open). $\endgroup$ – Clement C. May 27 '16 at 11:34
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So, to me it looks like you've proven (or at least understood the proof) that $f$ is injective since you assume $f(a_{1})=f(a_{2})$ and subsequently prove that $a_{1}=a_{2}$

As for proving $g$ is surjective, you almost have it: you know that $g(f(a))=b, \forall b \in X$, so for any $b \in X$, you could pick some $c \in Y$ such that $f(a) = c$ and you'd be done.

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One issue with your current attempt is that it's not clearly apparent what the structure of your argument is: you use $a_1,a_2$ without specifying what they are, not clearly stating what you are trying to establish with them. (Below is a detailed argument showing the two items you are asked to proved.)

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Assume $g\circ f$ is bijective.

  • To prove $f$ is injective, you need to show that for any $a,b\in X$ such that $f(a)=f(b)$ you must have $a=b$. So take any $a,b\in X$ such that $f(a)=f(b)$.

    Write $c = f(a)=f(b)$. Then $$ g\circ f(a) = g(f(a)) = g(c) = g(f(b)) = g\circ f(b) $$ but since $g\circ f$ is injective by assumption, this implies $a=b$. Since $a,b$ were arbitrary, this shows injectivity of $f$.

  • To prove $g$ is surjective, you need to show that for any $a\in X$ there is $c\in Y$ such that $g(c)=a$. So fix any $a\in X$.

    Since $g\circ f$ is surjective by assumption, this implies there is $b\in X$ such that $g\circ f(b)=a$. Let $c=f(b)\in Y$. Then $g(c) = g(f(b)) =a$. Since $a$ was arbitrary, this shows surjectivity of $g$.

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  • $\begingroup$ how can you infer the following statement in such a natural way: Let $c=f(b)\in Y$. Then $g(c) = g(f(b)) =a$. Can you elaborate on that? $\endgroup$ – TheMathNoob May 25 '16 at 6:21
  • $\begingroup$ @TheMathNoob The first equality is by definition of $c$, as $c=f(b)$; the second follows from $g(f(x)) = g\circ f(x)$ for every $x\in X$ (which is the definition of $g\circ f$) and the way we chose $b$ (i.e., $g\circ f(b)=a$). $\endgroup$ – Clement C. May 25 '16 at 6:23
  • $\begingroup$ yes thanks, now I have another question $\endgroup$ – TheMathNoob May 26 '16 at 2:05
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    $\begingroup$ @TheMathNoob Great. If this one is settled, consider accepting one of the two answers (checkmark on the left of either of them). $\endgroup$ – Clement C. May 26 '16 at 4:53

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