Abstract Algebra – Pushout of an Epimorphism is an Epimorphism

abstract-algebracategory-theoryepimorphisms

I'm reading the book "Handbook of Categorical Algebra: Volume 1, Basic Category Theory" by Francis Borceux, and in page 52 he states that

"…"the pullback of a monomorphism is a monomorphism". The dual notion of a "pullback" is that of a "pushout". In particular, "the pushout of an epimorphism is an epimorphism"…"

how can I show this last statement directly?

Best Answer

So suppose

$$\begin{array} AA & \stackrel{g}{\longrightarrow} & C \\ \downarrow{f} & & \downarrow{\beta} \\ B & \stackrel{\alpha}{\longrightarrow} & D \end{array}$$ is a pushout and $f$ is epi. We will show that $\beta$ is epi. Suppose, $h_1, h_2 \colon D \to D'$ are such that $h_1 \beta = h_2 \beta$. We have $$ h_1\alpha f = h_1 \beta g = h_2\beta g = h_2 \alpha f $$ As $f$ is epi, we have $h_1 \alpha = h_2 \alpha$. Now, as we have a pushout, there is a unique(!) $h \colon D \to D'$ such that $h\alpha = h_1\alpha = h_2\alpha$ and $h\beta = h_1 \beta = h_2 \beta$. As $h_1$ and $h_2$ both have this property, $h_1 = h_2$ and $\beta$ is epi.