Two mappings with constant composition Can you tell me if my answer is correct? Thank you so much!!!
Here is the problem:
If $f,g$ are mappings of $S$ into $S$ and $f\circ g$ is a constant function, then
(a) What can you say about $f$ if $g$ is onto?
(b) What can you say about $g$ if $f$ is 1-1.
Original Image
As $f\circ g$ is a constant function, $f\circ g$ is neither onto nor 1-1.
If $g$ is onto, $f$ cannot be onto because if $f$ is onto and $g$ is onto, $f\circ g$ would be onto.
If $f$ is $1-1$, $g$ cannot be 1-1 because if $f$ is 1-1 and $g$ is 1-1, $f\circ g$ would be 1-1.
In any case, the range of $g$ must be a subset of the domain of $f$.
 A: If $g$ is onto then you can write (for $y$ with $g(y)=x$):
$$f(x)=f(g(y))=\text{constant}$$
If $f$ is 1:1 then it has a left inverse $f^{-1}$. Thus:
$$g(x)=f^{-1}(f(g(x)))=f^{-1}(\text{constant})=\text{constant}$$
A: An informal, yet correct, answer provided you know what all the terms mean can be:
$g$ is onto means $g(S) = \{g(x)|x \in S\}$ can, and will be anything.  So $f(g(x))$ can, and will, have $g(x)$ being any value yet $f(g(x))$ is constant so if $g(x)=y$ is .... anything....$f(y)$ is constant.  So $f$ is constant.
$f$ is 1-1 means different input of $f$ will have different output.  So if $f(g(x)) = f(g(y)) = c$ is constant then the output of $f$ is the same for all possible input of $g(x)$ or $g(y)$and the only way that can happen is if the input is the same.  So all the $g(x)$ or $g(y)$ are the same no matter what $x$ or $y$ are.  So $g$ is constant.
It's informal, but those are the arguments and reasoning.
A: Hints/Solutions:
(a) $f$ is constant.
(b) $g$ is constant.
A: 1) If $g$ is onto then for any $x \in S$ there is a $y \in S$ so that $g(y) = x$.
So for every $x \in S$ there is some $y \in S$ so that $f(x) = f(g(y))$.  So what does that tell us about $f$?
2) If $f$ is 1-1 then for any $x \ne y$ we know $f(x) \ne f(y)$ so what can we say about $f(g(x))$ and $f(g(y))$ are those ever equal?  Never equal? What does that tell us about $g(x)$ and $g(y)$?
