1
$\begingroup$

Monomorhism as defined is: A morphism $f: A \to B$ is a monomorphism if for every object $C$ and every pair of morphisms $g,h: C \to A$ the condition $f\circ g = f\circ h$ implies $g = h$.

Applying this to sets, I could not think of counter example where $f\circ g = f\circ h$, but $g\neq h$. What are some examples of this case?

$\endgroup$

2 Answers 2

1
$\begingroup$

Let $A=\{1,2\}$, $B=C=\{1\}$, $g(1)=1$, $h(1)=2$, and $f(1)=f(2)=1$. Then $f\circ g=f\circ h$ but $g\neq h$. More generally, if $f$ is a constant function, then $f\circ g=f\circ h$ will always be true, no matter what functions $g,h:C\to A$ you have.

Even more generally, if $f$ is any non-injective function, you can find $g$ and $h$ which give a counterexample. I'll leave that as a challenge for you to prove; the idea is similar to the example above, where $f$ failed to be injective because $f(1)=f(2)$.

$\endgroup$
1
  • $\begingroup$ Thanks, is the following the proof: f(x) is non-injective => for some x and y, f(a) = c and f(b) = c, there can be a g and h such that g(x) = a and h(x) = b. Thus f o g = f o h = c, but g != h, $\endgroup$
    – coder_bro
    Nov 23, 2017 at 6:38
1
$\begingroup$

In $\mathsf{Set}$ we have that a morphism $f$ is a monomorphism if and only if $f$ is injective. So take any map which is not injective to construct appropriate maps $g$ and $k$. For example the mapping $f : x\mapsto x^2$ on the reals and $g = 1$ and $h= -1$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .