Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In $\mathbf{Top}$, if $A$ is a subspace of $X$ and $f: A \to Y$ is a continuous mapping, then $Y$ embeds into $X \cup_f Y$. I wonder if this generalizes to an arbitrary category.

Consider the pushout of a cospan $Y \leftarrow A \hookrightarrow X$, where the latter arrow is a regular monomorphism. Is the arrow parallel to the regular monomorphism itself a regular monomorphism?

share|improve this question
    
$X \cup_f Y$ is called a pushout, not a pushforward. –  Zhen Lin Dec 2 '12 at 14:35
1  
@ZhenLin OMG, I spent two years reading it as "pushforward" :( –  Alexei Averchenko Dec 2 '12 at 14:41
2  
After dualizing your question is equivalent to: "is the pullback of a regular epimorphism a regular epimorphism?". The answer is positive for regular categories. And this make me suspect that we can construct a counterexample for the general situation. However, I failed in finding one, for the moment. –  Mauro Porta Dec 4 '12 at 21:59
add comment

1 Answer 1

up vote 3 down vote accepted

The claim fails in $\textbf{Top}^\textrm{op}$, amusingly. Let $A = \{ a, b, c, d \}$ with open sets $$\emptyset, \{ a \}, \{ b \}, \{ a, b \}, \{ a, c \}, \{ b, d \}, \{ a, b, c \}, \{ a, b ,d \}, \{ a, b, c, d \}$$ let $B = \{ 0, 1, 2 \}$ with open sets $$\emptyset, \{ 0 \}, \{ 0, 1 \}, \{ 0, 1, 2 \}$$ and let $C = \{ 0, 2 \}$ be topologised as a subspace of $B$. Let $p : A \to B$ be the continuous map given by $$p(a) = 0, p(b) = 1, p(c) = 1, p(d) = 2$$ and let $i : C \to B$ be the inclusion. It is not hard to see that $p$ is a quotient map, so it is a regular epimorphism; now consider the pullback of $p$ along $i$. This is the map $q : D \to C$ where $D = \{ a, d \}$ is a discrete subspace of $A$, and $q$ is certainly surjective (so is an epimorphism) but $q$ is not a quotient map (so not a regular epimorphism). Thus, pullbacks in $\textbf{Top}$ do not preserve regular epimorphisms; equivalently, pushouts in $\textbf{Top}^\textrm{op}$ do not preserve regular monomorphisms.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.