Proving Epimorphism: Understanding Surjective Morphisms

[jhendren]

How would one show a morphism is an epimorphism iff it is surjective (ONTO)?

 

[Office_Shredder]

What is your definition of epimorphism if it's not that it is a surjective map?

 

[micromass]

What is your definition of epimorphism if it's not that it is a surjective map?

An epimorphism in category theory is a morphism $f:X\rightarrow Y$ such that if $g,h:Y\rightarrow Z$ are morphisms such that $g\circ f = h\circ f$, then $g=h$.

It is certainly not always true that an epimorphism is surjective. It depends on what category you work in. So, to the OP, you are talking about an epimorphism between which structures? Sets? Groups?

 

[PSarkar]

Must be on sets. Here is a Wikipedia page on it: https://en.wikipedia.org/wiki/Surjective_function#Surjections_as_epimorphisms

 

[blue_raver22]

To prove that a morphism is an epimorphism if and only if it is surjective, we must first understand the definitions of these terms. A morphism is a mathematical function that preserves the structure of a mathematical object, while an epimorphism is a special type of morphism that is "onto," meaning it maps every element in the target set to at least one element in the source set. In other words, an epimorphism is a surjective function.

To show that a morphism is an epimorphism if and only if it is surjective, we can use the definition of an epimorphism and the properties of surjective functions. Let f: X → Y be a morphism between two sets X and Y.

First, assume that f is an epimorphism. This means that for any element y in Y, there exists at least one element x in X such that f(x) = y. In other words, f is onto or surjective.

Conversely, assume that f is surjective. This means that for every element y in Y, there exists at least one element x in X such that f(x) = y. To show that f is an epimorphism, we need to prove that for any other morphism g: Y → Z, if f o g = f, then g = idY, the identity function on Y.

Since f is surjective, for every element z in Z, there exists at least one element y in Y such that f(y) = z. Therefore, for any element z in Z, there exists at least one element x in X such that f(g(x)) = z. Since f o g = f, we can substitute f(g(x)) for z, giving us f(g(x)) = f(y). By the definition of a function, if f(g(x)) = f(y), then g(x) = y. This means that for every element y in Y, g(x) = y, which implies that g = idY.

Therefore, we have shown that if f is surjective, then it is also an epimorphism. And if f is an epimorphism, then it is also surjective. Hence, a morphism is an epimorphism if and only if it is surjective.