Got a question as the following, could solve the first, would appreciate getting some help for the second :
Let there be $F(h)=h \circ \phi$, $F:C^B \to C^A$, $\phi : A\to B$. Assuming there's in C more then one element, prove:
- if F is injective then $\phi$ is surjective.
- if F is surjective then $\phi$ is injective.
So for "if F is injective then $\phi$ is surjective", I proved by contrapositivity. I assumed that F is injective but $\phi$ isn't surjective, then exists $b \in B$ such that there is no $a\in A$ that fulfills $\phi(a)=b$. Secondly, I define $f_1, f_2 \in C^B$ that are equal except for element 'b' ($\in B$), meaning that: $f_1(b)=c_1$ $f_2(b)=c_2$. Now, because F is injective $F(f_1)=F(f_2)$, then $f_1 = f_2$ though $f_1(b)\neq f_2(b)$. Then we get contradiction, then $\phi$ must be surjective.
Now, about "If F is surjective then $\phi$ is injective", I'm not quite sure how to do that. I also think it should be done by the contrapositive path. Should I define some function $g\in C^A$ that there is no $h\in C^B$ that gives $F(h)=g$ because of defining $\phi$ as non injective?
Would appreciate your help.
First, make $f_1$ and $f_2$ completely concrete: indeed we pick distinct $c_1 \neq c_2$ in $C$ (which our assumption tells us we can do). We have one unreachable $b \in B$. Define $f_1(x) = c_1 = f_2(x)$ for all $x \in B\setminus\{b\}$, $f_1(b) = c_1, f_2(b) = c_2$. Now explicitly show that $F(f_1) = F(f_2)$, etc.
Now suppose that $\phi$ is not injective, this gives us positive info: there are $a_1 \neq a_2$ in $A$ such that $\phi(a_1) = b = \phi(a_2)$ for some $b \in B$.
This means that all compositions with $\phi$ (which is what $F$ is) will do the same thing to $a_1$ and $a_2$. So this limits the kind of functions we can get as an image. Find a concrete (!) function that does not do this and then it cannot be an image of $F$ for this $\phi$.