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.

Answer 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 $$g(x)\ne f(x)$$

Answer 2

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

Answer 3

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 $A\neq B$.