Find example where: h ◦ f = f ◦ k, but: h ≠ k for a two element Set

Question

How do you prove that the following statement is incorrect:

If $h \circ f = f \circ k$, then $h = k$

When using the following guides:

• Use a set with two elements
• With three endomaps for $f$, $h$, and $k$
• Where $f$ is invertible

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I can prove the statement false (without the guides above) with the following scenario:

• $A = \{x, y, z\}$
• $f$ is a constant map to $x$
• $h$ maps $x$ to $x$, and the others to $z$
• $k$ is a constant map to $z$

In this scenario:

• $h \circ f$ is a constant map $x$
• $f \circ k$ is also a constant map $x$
• but $h ≠ k$

However the book I am following (Conceptual Mathematics: A First Introduction to Categories) specifically asks for a two element set with three endomaps.

Thanks for you help!

Answer 1

Let $A=\{1,2\}$, $f(1)=2,f(2)=1$, $h(1)=h(2)=1$ and $k(1)=k(2)=2$. Then $h\circ f=h$ and $f\circ k=h$ but $h\not=k$.

Answer 2

How about this? $$A=\{0,1\}$$ and $h(x)=0$, $k(x)=1$ and $f(x)=1-x$, $\forall x\in A$. Your idea of taking constant functions was good, you were close...