Are there any rules that apply to $f\circ g$ or $g\circ f$
if I know that $f$ is surjective but not injective,
and $g$ is also surjective but not injective?
Surjectivity or Injectivity of combination of functions
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functions
discrete-mathematics
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0Related: http://math.stackexchange.com/questions/75880/if-g-circ-f-is-the-identity-function-then-which-of-f-and-g-is-onto-and-w/ – 2011-12-01
1 Answers
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Assuming $f\colon A\to B$ and $g\colon B\to C$:
- If $f$ and $g$ are one-to-one, then $g\circ f$ is one-to-one.
- If $g\circ f$ is one-to-one, then $f$ is one-to-one and $g|_{f(A)}$ (the restriction of $g$ to the image of $f$) is one-to-one. But $g$ need not be one-to-one.
- Equivalently, if $f$ is not one-to-one, then $g\circ f$ is not one-to-one.
- If $f$ and $g$ are onto, then $g\circ f$ is onto.
- If $g\circ f$ is onto, then $g$ is onto, but $f$ need not be onto.
- Equivalently, if $g$ is not onto, then $g\circ f$ is not onto.
- If $f$ and $g$ are bijective, then $g\circ f$ is bijective.
- If $g\circ f$ is bijective, then $f$ is one-to-one, $g$ is onto, and $g|_{f(A)}$ is one-to-one, but that's the best you can say.
A standard example: take $A=\{1\}$, $B=\{x,y\}$, $C=\{2\}$. Take $f(1)=x$, $g(x)=g(y)=2$. Then $f$ is one-to-one and not onto; $g$ is onto and not one-to-one; and $g\circ f$ is bijective. $\begin{array}{ccccc} & & y & &\\ & & &\searrow\\ 1 & \stackrel{f}{\rightarrow} & x & \stackrel{g}{\rightarrow}& 2 \end{array}$
If $g$ is onto and $f$ is not, then $g\circ f$ may or may not be onto. If $f$ is one-to-one and $g$ is not, then $g\circ f$ may or may not be one-to-one. I'll let you construct examples of each.