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To clarify, consider in a category $\mathbf{C}$, a object $B$ and a terminal object $T$ both in Obj$_\mathbf{C}$, and a monomorphism $T \stackrel{f}{\rightarrowtail} B$. If $g$ in Mor$_\mathbf{C}$ is any other monomorphism with $B$ as codomain, lets say $A\stackrel{g}{\rightarrowtail} B $, such that $f \sim_\text{mono} g$, that is, there are morphisms $T\stackrel{h}{\rightarrow} A$ and $A\stackrel{s}{\rightarrow}T$ ($s \equiv \tau_A $) with $f=g \circ h$ and $g=f \circ s $, in a diagram:

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Then we can clearly get a morphism from any object $C$ in the category $\mathbf{C}$ to $A$ by way of $h\circ \tau_C$, but how does one prove that $h$ is unique making $A$ also a terminal object? (do I haveto use that $h$ is also a monomorphism since it has $T$ as a domain?)

Felipe Dilho
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1 Answers1

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Both $\tau_A$ and $h$ are monic, because the compositions $f\tau_A$ and $gh$ are monic. But the equations $f\tau_A h=f$ and $gh\tau_A$ imply (by monicity of $f$ and $g$) that $\tau_Ah=\mathrm{id}_T$ (this part actually also just follows from $T$ being terminal) and $h\tau_A=\mathrm{id}_A$. Therefore, $h\colon T\to A$ is an isomorphism, making $A$ a terminal object, and $h$ is the unique morphism $T\to A$.

Daniël Apol
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  • Why can we say from the outset that $f \circ \tau_A$ and $g \circ h$ are monic? If I try to symplify $(f \circ \tau_A) \circ a = (f \circ \tau_A) \circ b$ I only get to $\tau_A\circ a =\tau_A \circ b$ but not to $a=b$. – Felipe Dilho Jan 30 '24 at 05:10
  • Actually that one I can get from $T$ being terminal, but from where do I get that $g\circ h$ is monic? I could only get to $h \circ a = h \circ b$ and not to $a=b$, since $h$ is not necessarily monic. – Felipe Dilho Jan 30 '24 at 05:21
  • $f\tau_A=g$ and $gh=f$ are monic by assumption. – Daniël Apol Jan 30 '24 at 10:20