Proofs regarding composition of functions

Question

I'm having trouble approaching the following question: Is the following statement true or false, provide a proof or a counterexample.

If $h: A\rightarrow B, \ g: B\rightarrow C, \ f: B\rightarrow C$ are three functions and $ g\circ h=f\circ h $ then $ f = g$.

I have a feeling that it may be false but I am having trouble finding a counterexample.

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

1

Hint: If $f(a)=g(a)$ and $h(x)=a$ then ...

Explicit example:

Let $$g(x)=x \text{ and } f(x)=x^2$$ If $$h(x)=0$$ then $$g\circ h(x)=0=f\circ h(x)$$ but $$g(x)\ne f(x)$$

edited Jun 12 '20 at 10:38

Community

1

answered Dec 14 '13 at 04:35

E.O.

6,942

Thanks! I also figured out another counterexample where g(x) = x, f(x) = |x| and h(x) = x^2. – user115462 Dec 14 '13 at 04:48
@user115462 indeed that would work too – E.O. Dec 14 '13 at 04:49

score 0 · Answer 2 · answered Dec 14 '13 at 04:17

0

If $h$ takes two different inputs to the same output

answered Dec 14 '13 at 04:17

Ross Millikan

374,822

score 0 · Answer 3 · answered Dec 14 '13 at 06:23

Another source of counter examples is matrix multiplication. Since matrix multiplication is just the composition of linear functions over a vector space.

Consider,

$$ A = \left[ \begin{array} \ 0 & 1 \\ 1 &1\end{array} \right]$$

$$ B = \left[ \begin{array} \ 0 & 1 \\ 0 &1\end{array} \right]$$

$$ C = \left[ \begin{array} \ 0 & 0 \\ 0 &1\end{array} \right]$$

$$ AC = \left[ \begin{array} \ 0 & 1 \\ 1 &1\end{array} \right]\left[ \begin{array} \ 0 & 0 \\ 0 &1\end{array} \right] = \left[ \begin{array} \ 0 & 1 \\ 0 &1\end{array} \right]$$

$$ BC = \left[ \begin{array} \ 0 & 1 \\ 0 &1\end{array} \right]\left[ \begin{array} \ 0 & 0 \\ 0 &1\end{array} \right] = \left[ \begin{array} \ 0 & 1 \\ 0 &1\end{array} \right]$$

$$ AC = BC$$

So we have that the composition of $A$ with $C$ is the same as the composition of $B$ with $C$ even though $A\neq B$.

Proofs regarding composition of functions

3 Answers3