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Revision difference : strong monomorphism
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Let $\mathcal{C}$ be a category. A monomorphism $f:A\to B$ in $\mathcal{C}$ is said to be a \emph{strong monomorphism} if, whenever we are given the following commutative diagram Let $\mathcal{C}$ be a category. A monomorphism $f:A\to B$ in $\mathcal{C}$ is said to be a \emph{strong monomorphism} if, whenever we are given the following commutative diagram
$$\xymatrix@+=4pc{ $$\xymatrix@+=4pc{
{C}\ar[r]^{g}\ar[d]_{x}&{D}\ar[d]^{y}\\ {C}\ar[r]^{g}\ar[d]_{x}&{D}\ar[d]^{y}\\
{A}\ar[r]_{f}&{B} {A}\ar[r]_{f}&{B}
} }
$$ $$
with $g$ an epimorphism, then there is a morphism $h: D\to A$ such that the following is another commutative diagram: with $g$ an epimorphism, then there is a morphism $h: D\to A$ such that the following is another commutative diagram:
$$\xymatrix@+=4pc{ $$\xymatrix@+=4pc{
{C}\ar[r]^{g}\ar[d]_{x}&{D}\ar[d]^{y} \ar@{.>}[dl]_{h} \\ {C}\ar[r]^{g}\ar[d]_{x}&{D}\ar[d]^{y} \ar@{.>}[dl]_{h} \\
{A}\ar[r]_{f}&{B} {A}\ar[r]_{f}&{B}
} }
$$ $$
Note that the ``diagonal'' morphism $h$ is necessarily unique. Note that the ``diagonal'' morphism $h$ is necessarily unique.
Dually, a \emph{strong epimorphism} is an epimorphism $f:A\to B$ such that, given a commutative diagram Dually, a \emph{strong epimorphism} is an epimorphism $f:A\to B$ such that, given a commutative diagram
$$\xymatrix@+=4pc{ $$\xymatrix@+=4pc{
{A}\ar[r]^{f}\ar[d]_{x}&{B}\ar[d]^{y}\\ {A}\ar[r]^{f}\ar[d]_{x}&{B}\ar[d]^{y}\\
{C}\ar[r]_{g}&{D} {C}\ar[r]_{g}&{D}
} }
$$ $$
with $g$ a monomorphism, then there is a (unique) morphism $h:B\to C$ such that the following diagram is again commutative: with $g$ a monomorphism, then there is a (unique) morphism $h:B\to C$ such that the following diagram is again commutative:
$$\xymatrix@+=4pc{ $$\xymatrix@+=4pc{
{A}\ar[r]^{f}\ar[d]_{x}&{B}\ar[d]^{y}\ar@{.>}[dl]_{h}\\ {A}\ar[r]^{f}\ar[d]_{x}&{B}\ar[d]^{y}\ar@{.>}[dl]_{h}\\
{C}\ar[r]_{g}&{D} {C}\ar[r]_{g}&{D}
} }
$$ $$
\textbf{Remark}. Every regular monomorphism is strong (see proof \PMlinkname{here}{RegularMonomorphism}), and every strong monomorphism is \PMlinkname{extremal}{ExtremalMonomorphism}. \textbf{Remark}. Every regular monomorphism is strong (see proof \PMlinkname{here}{RegularMonomorphism}), and every strong monomorphism is \PMlinkname{extremal}{ExtremalMonomorphism}.
\begin{proof} \begin{proof}
Suppose $f:A\to B$ is a strong monomorphism and that $f=h\circ g$ with $g:A\to C$ epimorphic. Then we have the following commutative diagram Suppose $f:A\to B$ is a strong monomorphism and that $f=h\circ g$ with $g:A\to C$ epimorphic. Then we have the following commutative diagram
$$\xymatrix@+=3pc{ $$\xymatrix@+=3pc{
{A}\ar[r]^{g}\ar[d]_{1_A}&{C}\ar[d]^{h}\\ {A}\ar[r]^{g}\ar[d]_{1_A}&{C}\ar[d]^{h}\\
{A}\ar[r]_{f}&{B} {A}\ar[r]_{f}&{B}
} }
$$ $$
Since $f$ is strong, there is a morphism $e:C\to A$ such that the diagram below is commutative Since $f$ is strong, there is a morphism $e:C\to A$ such that the diagram below is commutative
$$\xymatrix@+=3pc{ $$\xymatrix@+=3pc{
{A}\ar[r]^{g}\ar[d]_{1_A}&{C}\ar[d]^{h} \ar@{.>}[dl]_{e} \\ {A}\ar[r]^{g}\ar[d]_{1_A}&{C}\ar[d]^{h} \ar@{.>}[dl]_{e} \\
{A}\ar[r]_{f}&{B} {A}\ar[r]_{f}&{B}
} }
$$ $$
This shows that $g$ is a split monomorphism, as $1_A=e\circ g$. But $g$ is epimorphic, we conclude that $g$ is an isomorphism (this fact is proved \PMlinkname{here}{PropertiesOfRegularAndExtremalMonomorphisms}). This shows that $g$ is a split monomorphism, as $1_A=e\circ g$. But $g$ is epimorphic, we conclude that $g$ is an isomorphism (this fact is proved \PMlinkname{here}{PropertiesOfRegularAndExtremalMonomorphisms}).
\end{proof} \end{proof}
\begin{thebibliography}{9}
\bibitem{fb} F. Borceux \emph{Basic Category Theory, Handbook of Categorical Algebra I}, Cambridge University Press, Cambridge, 1994
\end{thebibliography}