Injective, Surjective Question on functions

Question

Consider two functions $:→$ and $:→$

Decide whether each of the following statements is true or false, and prove each claim.

a) If $∘$ is injective, then $$ is injective.

b) If $∘$ is surjective, then $$ is surjective.

c) If $∘$ is surjective and $$ is injective, then is surjective.

For part a, injective means: $f(x)=f(y)→x=y$ and therefore is true.

I am unsure about part b and c. How do I prove and solve the 3 parts?

Answer 1

a) is true, take $f(a)=f(b)$ then $g(f(a)) = g(f(b))$ and since $g\circ f$ is injective we get $a=b$ so $f$ is injective.

b) is not true, take $f(x)=\arctan x$ and $g(x)=\tan x$

Then $g(f(x))=x$ so it is surjective while $f$ is not.

The following is wrong: (but it is usefull to see why, look at the comment):

c) is not true: Take $ f(x)=e^x$ and $g(x)=\ln x$. Then $g(f(x))=x$ is surjective and $g$ injective, while $f$ is not surjective.

Answer 2

a) true (but what you are saying is not a proof of it).

Let $s,s'\in S$ with $f(s)=f(s')$. Then also $(g\circ f)(s)=g(f(s))=g(f(s'))=(g\circ f)(s')$ so that we are allowed to conclude that $s=s'$ because $g\circ f$ is injective.

b) false in general.

Let $U=\{u\}$ and $S\neq\varnothing$ so that automatically $g\circ f$ is surjective. If however $T$ contains more than one element then we can find a function $f:S\to T$ that is not surjective.

c) true.

Let $t\in T$. Then $g(t)\in U$ and since $g\circ f$ is surjective we can find some $s\in S$ with $g(f(s))=g(t)$. Then the injectivity of $g$ tells us that $t=f(s)$. This proves that $f$ is surjective.