A surjective function is an epimorphism in the category Set

up:: Epimorphism

Let $f:X\to Y$ and $g_1,g_2:Y\to Z$ be functions in the Category $Set$, such that $g_1\circ f=g_2\circ f$. We seek to prove that $f$ is surjective $⟹$ $g_1=g_2$ under these conditions.

Surjective $⟹$ Epimorphism

Assume $f$ is Surjective. Then it’ll follow that

$$\forall y\in Y:\exists x\in X\mid y=f(x)$$

Thus, for any $y\in Y$, we have that

$$g_1(y)=g_1(f(x))\stackrel{Hyp.}{=}{g}_2(f(x))=g_2(y)$$

And thus $f$ is an Epimorphism.

Epimorphism $⟹$ Surjective

Proof by contrapositive: Suppose $f$ is not surjective. Then there are points in $Y\setminus f(X)$ ─ let $\tilde{y}\in Y\setminus f(X)$.

Then we can construct functions $g_1,g_2:Y\to Z$ such that

$$\{\begin{array}{l}\forall x\in X:g_1(f(x))=g_2(f(x))\\ g_1(\tilde{y})\ne g_2(\tilde{y})\end{array}$$

Thus, it would follow that $f$ is not an epimorphism, since there is some point in $Y$ such that they differ, regardless of their composition being equal.

Therefore, all epimorphisms $f$ are surjective (in the category $Set$).

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References

• epimorphism in nLab