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Let $f\colon A\to B$ and $g\colon B\to C$ be functions such that $g\circ f$ is surjective:

  1. Must $f$ be surjective, either prove or give a counterexample.
  2. Must $g$ be surjective, either prove or give a counterexample.

What I have so far: Let A = {1, 2} B = {3,4} and C = {5}

f : A --> B is defined as f(1) = 3 and f(2) = 3

g : B --> C is defined as g(3) = 5 and g(4) = 5 Since the entire codomain of C is covered, g o f is surjective but f is not surjective as it does not cover the entire codomain of B.

Alex
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mathacc
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  • Experiment with various very small sets $A,B,C$, say having size $1$ or $2$ (not necessarily all the same size). What can you discover? – Greg Martin Oct 23 '23 at 18:14
  • so like A = {100, 101], B = {200, 201} and C = {300, 301}? wouldnt they both be surjective in this case? – mathacc Oct 23 '23 at 18:22
  • In your example, you've not specified the functions. One case with all bijections is trivial, so try where your small sets have slightly different cardinalities. Draw simple pictures. – AlgTop1854 Oct 23 '23 at 18:32
  • Like $A={"dog", "cat"}, B={"red", "blue"}, C={"tree"}$ – Galois group Oct 23 '23 at 18:33
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    In my opinion, the only way to attack this problem is to first try to expand your intuition around the idea of a function being surjective. Consider any surjective function $~h:A \to B,~$ and let $~C~$ denote a superset to $~B,~$ such that $~B \neq C.~$ Then $~h:A \to C~$ is not surjective. – user2661923 Oct 23 '23 at 19:07
  • I think that functions f and g both must be surjective for their composition to be surjective. See thi at time stamp 7:54 https://youtu.be/aVd2rYElUh0?si=KmawQ52vCPtdmDVd – Popocatepetl112358 Oct 23 '23 at 23:04

2 Answers2

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So we are given $f: A \rightarrow B$ and $g : B \rightarrow C$. And so we know that $$g \circ f : A \rightarrow C$$ and are given that $g \circ f$ is surjective. Remember a funtion $$h : D \rightarrow C$$ is surjective if the range of the function is equal to the codomain of the function.

Hints

From this what can we conclude about $g$ if it also maps to $C$?

Can we conclude anything about $B$? And so can we conclude anything about $f$ being surjective?

Also as mentioned in the comments, using random examples can be a good way to get a better idea of the problem, especially as you may need to find a counterexample. For example we can consider \begin{align} f : \{1,2\} &\rightarrow \{1,2,3\} \\ 1 &\mapsto 1 \\ 2 &\mapsto 2 \end{align}

\begin{align} g:\{1,2,3\} &\rightarrow \{4\} \\ x &\mapsto 4\end{align} Now what is the range and codomain of $f$ and are they equal?

Jacob Walker
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  • There is a proposition that says that if functions f and g are injective (surjective), then their composition must be injective (surjective). Though I do not know if thr converse is true or if one of f or g are surjective, then their composition would be surjective. See this starting at 7:54 https://youtu.be/aVd2rYElUh0?si=KmawQ52vCPtdmDVd – Popocatepetl112358 Oct 23 '23 at 20:21
  • f can not be surjective by counter example, but Im confusing myself on how I would apply that, for f not to be surjective means that there has to be an element in the domain of b that is not the domain of a, so if the example was A = {2} B = {2, 3} and C = {4}. f is not surjective as 3 is not an element in the domain of A? – mathacc Oct 23 '23 at 20:40
  • @mathacc you seem to be confusing your domain, codomain and range. Surjectivity has nothing to do with the domain. Also with your example you need to define the function. To define a function we the domain and codomain as well as a map between the two. I have edited my answer with an example for you to consider. – Jacob Walker Oct 23 '23 at 21:57
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If $g \circ f$ is surjective, then $g$ must also be surjective, but not necessarily $f$.

Given any $c \in C$, there is some $a \in A$ for which $g(f(a))=c$ by surjectivity of $g \circ f$. This means that $g(b)=c$ for $b=f(a)$, so $g$ must also be surjective.

If $g$ is a surjective non-injective function, then let $b_1$ and $b_2$ be two distinct elements of $B$ for which $g(b_1)=g(b_2)$. Then, if $A=B \setminus \{b_1\}$ and $f$ is the inclusion map, $g \circ f=g|_{A \setminus \{b_1\}}$ is still surjective, but $f$ is not surjective.

Geoffrey Trang
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