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How would you prove the following facts:

(a) If $f\colon A \to B$ and $g \colon B \to C$ are injective then $g \circ f$ is injective.
(b) If $f\colon A \to B$ and $g \colon B \to C$ are surjective then $g \circ f$ is surjective.
(c) If $f\colon A \to B$ and $g \colon B \to C$ are bijective then $g \circ f$ is bijective.

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marked as duplicate by Jonas Meyer, Ivo Terek, Mike Miller, JimmyK4542, Behaviour Dec 28 '14 at 4:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I don't know how to solve it so I'm more of asking for help doing it. From my researches I found f(x1)=f(x2), gof(x1)=g(f(x1))=g(f(x2))=gof(x2) => x1=x2 – Max Nov 15 '12 at 11:46
I would begin by verifying that the definitions for injective, surjective and bijective holds. – Asaf Karagila Nov 15 '12 at 11:49
I'll do the first one. If $g(f(x_1))=g(f(x_2))$ then $f(x_1)=f(x_2)$ since $g$ is injective, so $x_1=x_2$ since $f$ is injective. Now can you do the others? – wj32 Nov 15 '12 at 11:51
@Max: Are you here to learn how to prove these claims, or are you here to get a ready-made answer for your homework sheet? Look at my answer, look at the comments. You should understand the definitions and from there it is really not that hard. – Asaf Karagila Nov 15 '12 at 12:09
What you wrote in that last comment makes no sense. If you have a book, open the book. If you have notes, open the notes. Use the definitions you are given. Don't think you can immediately remember everything you were told. You can't. You have to use things before you can remember them, and it is obvious that you haven't really used the definitions of surjection, injection, and so on. – Asaf Karagila Nov 15 '12 at 14:03

What you need to do is to understand the definitions of injective, surjective and bijective, as well as the definition of composition. Then it is just a matter of unwinding the definitions until you completed the proof.

Let me help you with one of those. Recall that $f$ is injective if $f(a)=f(b)\iff a=b$. Recall also that $g\circ f\colon A\to C$ is defined by $(g\circ f)(a)=g(f(a))$.

Now suppose that $f,g$ are both injective. To show that $g\circ f$ is injective we want to show that if $(g\circ f)(a)=(g\circ f)(a')$ then $a=a'$.

By the definition of composition $(g\circ f)(a)=g(f(a))=g(f(a'))=(g\circ f)(a')$. However $g$ is injective so we know that $f(a)=f(a')$. Since $f$ is injective we know that $a=a'$, as wanted.

The idea is very similar for the other parts. Just understand the definitions and what you have to show.

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Proving that the composite of injections is an injection

$g \circ f(x)=g \circ f(y)\Rightarrow g(f(x))=g(f(y)) \Rightarrow f(x)=f(y) \Rightarrow x=y$ ($f$ and $g$ are injective, $f^{-1}$ and $g^{-1}$ exists)

Proving that the composite of surjections is a surjection

$\forall w \in C: \exists y \in B: g \left({y}\right)=w$ $\Rightarrow \exists x \in A: f \left({x}\right)=y$

We have used the definitions of $f$ and $g$ being surjective.

$g \circ f(x)=g(f(x))=g(y)=w.$

Hence $g \circ f$ is a surjective.

Concerning the last part, a function being bijective means that it is both injective and surjective, hence the result follows.

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