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The question asks:

Let $g : A \to B$ and $f : B \to C$ be two functions.

Show that if $f \circ g$ is a bijection, then $f$ is a surjection and $g$ is an injection

I know how to prove if it's given that either $f$ or $g$ is either injective or surjective, but I am not quite sure how to approach this question since it's asking for both.

Also, how do I show that $f \circ g$ is a bijection?

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First, we note that $f \circ g$ will be a map $A \to C$. If we assume that $f \circ g$ is bijective, we automatically have the following two observations:

  1. Surjectivity: for each $c \in C$, there exists $a \in A$ such that $(f \circ g)(a) = c$.
  2. Injectivity: if $(f \circ g)(a) = (f \circ g)(a^\prime)$, then we must have $a = a^\prime$.

All that is needed from here are the two properties above. I'll write out the injectivity part and leave the surjective part to you.

To check that $g$ is injective, we proceed directly using the definition. Let's assume that $g(a) = g(a^\prime)$ for two $a,a^\prime \in A$. Applying $f$ to both sides of this equation, we find that $$ (f \circ g)(a) = f(g(a)) = f(g(a^\prime)) = (f \circ g)(a^\prime) $$ whence we must have $a = a^\prime$ (why? which property is used here?). Using 1. in a similar way, we can show that $f$ must be surjective as well.

Also, the hypothesis is that $f \circ g$ is bijective. Consequently, you do not need to prove that $f \circ g$ is bijective as this property is given!

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Use the fact that $f\circ g$ is a surjection to prove that $f$ is a surjection, too. As a great big hint: $(f\circ g)(x)=f\bigl(g(x)\bigr).$

Use the fact that $f\circ g$ is an injection to prove that $g$ is an injection, too. As a great big hint: $(f\circ g)(x)=f\bigl(g(x)\bigr)$ and $(f\circ g)(y)=f\bigl(g(y)\bigr).$

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  • $\begingroup$ how about proving f◦ g is a bijection? $\endgroup$ – ph-quiett Apr 7 at 19:04
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    $\begingroup$ That isn't what you've been asked to do. Rather, you're supposed to assume that $f\circ g$ is a bijection, and use that to prove facts about $f$ and $g.$ $\endgroup$ – Cameron Buie Apr 7 at 19:18
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Hint:

Prove these more general assertions:

  1. If $f\circ g$ is injective, then $g$ is injective (by contrapositive).
  2. If $f\circ g$ is surjective, then $f$ is surjective (observe that $(f\circ g)(A)\subset g(B)$).
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    $\begingroup$ There's really no need for contrapositive, since $f$ is a function. $\endgroup$ – Cameron Buie Apr 7 at 19:19
  • $\begingroup$ Injectivity is really very short by contrapositive. $\endgroup$ – Bernard Apr 7 at 19:22
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    $\begingroup$ True. The direct proof is shorter, though. $\endgroup$ – Cameron Buie Apr 7 at 19:25
  • $\begingroup$ Shorter than this: ‘If $g$ is not injective, obviously $f\circ g$ can't be either.’? $\endgroup$ – Bernard Apr 7 at 19:40
  • $\begingroup$ I would contend that use of the word "obviously" would be ill-advised at this level of mathematics, so wouldn't constitute an acceptable proof, unless followed up by "because [justification from the definition]." $\endgroup$ – Cameron Buie Apr 7 at 19:48

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