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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?

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1  
If $f$ and $g$ are both surjective, then so are $f \circ g$ and $g \circ f$ (of course, the later one is redundant). –  Matthias Klupsch Nov 20 '11 at 19:49
    
Moreover, if $g$ is not injective, then $f \circ g$ is not injective. –  Joel Cohen Nov 21 '11 at 2:16
    
and if only f is surjective? does it necessarily mean that f∘g is not surjective ? –  Asaf Nov 21 '11 at 11:57
    
@Asaf: You can have $f\circ g$ surjective with $f$ surjective and $g$ not surjective. –  Arturo Magidin Nov 21 '11 at 14:38
    

1 Answer 1

up vote 1 down vote accepted

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.

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