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Let $$f : A \to B, \quad g: B \to C, $$ be two functions. Show the following:

1) If $f$ and $g$ are surjective then $g \circ f$ is surjective

2) If $f$ and $g$ are bijective then $g \circ f$ is bijective.

3) If $g\circ f$ is injective then $f$ is injective.

4) If $g \circ f$ is surjective then $g$ is surjective.

5) If $g \circ f$ is bijective then $f$ is injective and $g$ is surjective.

How can I prove the following statements ? I assume that I can say that 2) is bijective If I can prove 1) (I know $g\circ f$ is injective)

A little bit help would be awesome thank you...


marked as duplicate by Arnaud D., The Phenotype, GNUSupporter 8964民主女神 地下教會, Strants, Xam Mar 1 '18 at 19:37

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  • 1
    $\begingroup$ What have you tried so far? Do you know the definitions of surjective, injective bijective? Do you understand these definitions? $\endgroup$ – Cettt Mar 1 '18 at 13:00
  • $\begingroup$ Welcome to stackexchange. You are more likely to get answers rather than downvotes or votes to close if you edit the question to show us just what you tried and where you are stuck, in as much detail as possible. Please ask just one question per post. $\endgroup$ – Ethan Bolker Mar 1 '18 at 13:01

The first and most important, you might keep some basic examples in mind.

Like the sets with different size. An easy way to consider these problems is taking some sets with different size for example.You can image this in the cases such as finite discrete set ($A = \{1,2,3,4,5\}$) or infinitely countable, you can see it as some vector spaces if you want.

Rough speaking, if $f : A \rightarrow B $ is surjective, you may think $B$ is "smaller" than $A$. On the other hand, if $f$ is injective, you may think $B$ is "bigger" than $A$. Of course, if $f$ is bijective, then these two guys will have the same size.

Your thought will be right because "Bijective" is stronger than "Surjective". Recall that bijection means both surjection and injection.

I hope these ideas will work for you.


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