$ \def\Vec#1{\mathbf{#1}} \def\vt#1{\Vec{v}_{#1}(t)} \def\v#1{\Vec{v}_{#1}} \def\vx#1{\Vec{x}_{#1}} \def\av{\bar{\Vec{v}}} \def\vdel{\Vec{\Delta}} $

Harald Kirsch

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In Category Theory, a monomorphism is a morphism $f:X\to Y$ for which the following condition holds: For any two morphisms $g_1, g_2: Z\to X$ the identity $f\circ g_1 = g\circ g_2$ induces $g_1 = g_2$.

As an Injective Function

Monomorphisms are the categorial way to describe an injective function, in that an injective function on sets is always a monomorphism.

Let $f:X\to Y$ be an injective function for the sets $X$ and $Y$. This in particular means that there is a function $h:f(X)\to X$ such that $h(f(x)) = x$ for all $x\in X$.

Now lets have another set, $Z$, and two functions $g_1, g_2: Z\to X$ such that $f\circ g_1 = f\circ g_2$ which means in particular that $$f(g_1(z)) = f(g_2(z))$$ for all $z\in Z$. Therefore we get for each $z$

$$\begin{align*} g_1(z) &= h(f(g_1(z)))\\ &= h(f(g_2(z))) \\ &= g_2(z)\,. \end{align*}$$

This means that $g_1=g_2$, and hence that $f$ is a monomorphism.

To Note

The categorial way to define a monomorphism cannot talk about the elements of the category's objects, because they have none, but it nevertheless manages to define an injective function.