1

Let $f : A \mapsto B$ and $g : B' \mapsto C$ where $B'$ is the range of $f$. Give a proof or a counter-example of the following.

(a) $\text{If }g \circ f \text{ is injective, }f\text{ is injective.}$

(b) $\text{If }g \circ f \text{ is injective, }g\text{ is injective.}$

(c) $\text{If }g \circ f \text{ is surjective, }f\text{ is surjective.}$

(d) $\text{If }g \circ f \text{ is surjective, }g\text{ is surjective.}$

I know the definitions for when a function is injective, surjective, bijective, etc. However, I am not sure what the notation of $f : A \mapsto B$ and $g : B' \mapsto C$ where $B'$ is the range of $f$ means in this case, Wikipedia's tells me on its injective function that a function is injective if f(x) = f(y). How does function composition tie in to that? Possibly g(f(x)) = g(f(y))?

SchrodingersCat
  • 24,472
WensworthKoohli
  • 47
  • 6
  • $f : A \to B$ means $f$ takes elements from $A$ as input and produces elements of $B$ as output. It may not be the case that every element of $B$ is output by $f$. But the set of elements that are output by $f$ (aka the range of $f$) is denoted in this problem as $B'$. Note that $B' \subseteq B$. –  Jan 26 '17 at 15:23
  • 1
    "function is injective if $f(x)=f(y)$" doesn't make any sense... – user160738 Jan 26 '17 at 15:26
  • You could search on injective surjective on this site and will find many questions addressing the base question. One of them is here – Ross Millikan Jan 26 '17 at 15:31
  • 1
    Do read carefully. The function $f$ is injective if $f(x)=f(y)$ implies that $x=y$. – George Law Jan 26 '17 at 15:40

1 Answers1

0

When we write $f : A \mapsto B$ we don't guarantee that all of $B$ is in the range of $f$. If $f$ is surjective it is, but it may not be. If we want to compose $f$ with $g$ and get $g \circ f $ we only require that $g$ be defined on the points that are in the range of $f$. For example, if we define $f: \Bbb N \mapsto \Bbb N$ as $f(x)=2x$ the range of $f$ is only the even numbers. If we want to discuss $g \circ f$ we only need to insist that $g$ be defined on the evens. That is why they have the distinction between $B$ and $B'$. You have not quoted the definition of injective correctly, which is that $f(x)=f(y) \implies x=y$. Now if we want to ask if $g \circ f$ is injective, we are asking if $g( f(x))=g(f(y)) \implies x=y$

Ross Millikan
  • 374,822
  • So then $f(x) = f(y)$ for some $x, y \in A$ Gives $g(f(x)) = g(f(y))$ = $(g \circ f)(x) = (g \circ f)(y)$ Which makes $g \circ f$ injective, because $x = y$, correct? That satisfies (a) but how would I do (b)? (b) would need a counter example, because it is not true for all cases I think. – WensworthKoohli Jan 27 '17 at 00:02
  • No, if $f$ is not injective and $f(x)=f(y)$ then $g(f(x))=g(f(y))$ so $g \circ f$ is known not to be injective. That should help you find a counterexample. – Ross Millikan Jan 27 '17 at 00:09