An injective function is a monomorphism in the category Set

up:: Monomorphism

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

Injective $⟹$ Monomorphism

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

$$\forall z\in Z:f(g_1(z))=f(g_2(z))⟹\forall z\in Z:g_1(z)=g_2(z)⟹g_1=g_2$$

Thus, $f$ is a monomorphism.

Monomorphism $⟹$ Injective

Let $f$ be a Monomorphism. Then for any two functions $g_1,g_2:Z\to X$, it’ll follow that $(f\circ g_1=f\circ g_2)⟹g_1=g_2$, for any functions $g_1,g_2:Z\to X$.

The proof is “constructive”: Let $x_1,x_2\in X$ and some $\tilde{z}\in Z$. Then choose $g_1,g_2$ such that

$$\{\begin{array}{l}{g}_1(z)=g_2(z):z\ne \tilde{z}\\ g_1(\tilde{z})=x_1\\ g_2(\tilde{z})=x_2\end{array}$$

that is, they are the same function, but they (may) differ for some $\tilde{z}\in Z$.

Then we have per hypothesis that

$$\forall z\in Z:f(g_1(z))=f(g_2(z))⟹g_1(z)=g_2(z)$$

In particular, we have that it follows for $\tilde{z}$:

$$\begin{array}{rl}& f(g_1(\tilde{z}))=f(g_2(\tilde{z}))⟹g_1(\tilde{z})=g_2(\tilde{z})\\ & ⟺f(x_1)=f(x_2)⟹x_1=x_2\end{array}$$

Since this can be done for all points $x_1,x_2\in X$, then $f$ is injective¹.

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References

• monomorphism in nLab

Footnotes

1. And, in fact, this proves that, given a monomorphism, we cannot construct $g_1,g_2$ as above such that they differ in a single point: the monomorphism condition ensures that they are necessarily equal to all points ─ provided that their composition with $f$ is equal. ↩