Suppose $g:A\rightarrow B$ and $f:B\rightarrow C$, where $f \circ g$ is injective. (Con't in description)

Question

Remember that the "circle operator" is a point-wise application of $g$ to the results of $f$. Answer the following questions and state reasons for your answer:

a) Suppose $g$ is injective. Must $f$ therefore be injective as well? -Yes/No -Reason?

B) Suppose $f$ is injective. Must $g$ therefore be injective as well? -Yes/no -Reason?

**I am really struggling to understand this. If you could please explain the reasoning behind the answer in a way that makes it easy to understand, I will really appreciate the help! (Not that I don't already)

Answer 1

Let me answer (A) and leave (B) for you to think about: Suppose $f\circ g$ and $g$ are injective, the question is whether $f$ is injective. So suppose $x,y\in B$ such that $f(x) = f(y)$, can you conclude that $x=y$?

Well, if $x=g(a), y=g(b)$ for some $a,b\in A$, then $$ f(g(a)) = f(g(b)) $$ and since $f\circ g$ is injective, you could conclude that $a=b$ and so $x=y$.

Unfortunately, there is no guaranteeing that such an $a$ or $b$ may exist.

And this insight allows you construct a counterexample:

Choose the domain of $f$ to be larger than the range of $g$, and define $f$ specifically so that it is not injective on the complement of the range of $g$.

For instance, take $$ A = \{1\}, B = \{1,2,3\}, C = \{0,1\} $$ Define $g:A\to B$ by $g(1) = 1$. Define $f:B\to C$ by $$ f(1) = 1, f(2) = f(3) = 0 $$ Now $g$ and $f\circ g$ are clearly injective, but $f$ is not.

Note that the earlier argument has actually proved something positive as well

If $f\circ g$ is injective and $g$ is surjective, then $f$ is injective.

Answer 2

a) $g$ injective means that two different points are never mapped together ($x\neq y \Rightarrow g(x)\neq g(y)$). But when you consider $(f\circ g)(x))=f(g(x))$, why can injectivity fail? $f$ could do anything, like mapping everything to one element. e.g. $f(x):=0$ for all $x$, then $f\circ g$ is constantly zero, no matter what properties $g$ has.

b) If $f$ is injective, think about what happens, if $g$ were not and it would map two different elements $x\neq y$ to the same image $g(x)=g(y)$. What consequence do you see for $f \circ g$? Well no matter what $f$ does, it has no chance to map one element to two different ones (because it is a map). Thus you would have $f(g(x))=f(g(y))$ and $f$ is not injective, which is a contradiction. Therefore $g$ has to be injective.