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Trying to understand Example on page 50 of David I. Spivak's book "Category Theory for Scientists," Spivak gives $X$ as the interval $[0..1]=\{x\in\mathbb{R}|\,0\le x\le 1\}$, $Y$ as the interval $[1..2]=\{x\in\mathbb{R}|\,1\le x\le 2\}$, and $W$ as the singleton set $\{1\}$. He then states that the pushout $X\overset{f}{\longleftarrow}W\overset{g}{\longrightarrow}$, where $f$ and $g$ are the "obvious" maps $(1_W\mapsto 1_X)$ and $(1_W\mapsto 1_Y)$, is $[0..2]=\{x\in\mathbb{R}|\,0\le x\le 2\}$, "as expected." But it looks to me like the pushout is supposed to be the quotient of $X\sqcup W\sqcup Y$ by an equivalence relation generated by the "obvious" functions, which would make the pushout a set with two elements, each of which is an equivalence class in $X\sqcup W\sqcup Y$; the first equivalence class being $1_W$, $1_X$, $1_Y$ and the second equivalence class being $[0..1)\cup(1..2]=\{x\in\mathbb{R}|\,0\le x\lt 1\}\cup\{x\in\mathbb{R}|\,1\lt x\le 2\}$. Where am I going wrong? Details of my reasoning follow.

I read the pushout as any set isomorphic to the fiber sum, which is the quotient set of the triple disjoint union $X\sqcup W\sqcup Y$ by the equivalence relation generated by the equivalences $w\, {\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}}f(w)$ and $w\, {\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}}g(w)$ (quoting loosely from the book, Definition, page 50). The equivalences, here, seem to need some interpretation.

Because equivalences are defined only in the cartesian product of a set with itself (definition, page 47), I interpret the ${\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}}$ relation to actually be in the cartesian product set $(X\sqcup W\sqcup Y)\times (X\sqcup W\sqcup Y)$; and the notation $w\, {\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}}f(w)$ to actually mean $w_W\, {\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}}f(w)_X$, where $w_W$ is the element $w$ in the disjoint union $X\sqcup W\sqcup Y$ that "remembers" it was included from set $W$, and $f(w)_X$ is the element $f(w)$ in the disjoint union $X\sqcup W\sqcup Y$ that "remembers" it was included from set $X$, both by the canonical inclusion maps from $W$ and $X$ into the disjoint union. Ditto for $w\, {\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}}g(w)$, which I take to actually mean $w_W\, {\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}}g(w)_Y$.

Under that interpretation (the only reasonable one I can see for the relation), I find that the three elements $1_W$, $1_X$, $1_Y$ form one equivalence class in $X\sqcup W\sqcup Y$, and that all other points, namely $[0..1)_X\cup(1..2]_Y=\{x_X|\,x\in\mathbb{R}|\,0\le x\lt 1\}\cup\{y_Y|\,y\in\mathbb{R},1\lt y\le 2\}$ form another equivalence class. Therefore, I see two elements in the fiber sum, i.e., the pushout. I can't find a way to make Spivak's claimed pushout $[0..2]=\{x\in\mathbb{R}|\,0\le x\le 2\}$ isomorphic to the fiber sum because the fiber sum puts $1$ in one equivalence class and all other points in another equivalence class.

What have I missed?

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up vote 3 down vote accepted

In the example in your question, the two "obvious" functions each contain one pair, so the relevant equivalence relation is generated by those two pairs, namely $(1_W,1_X)$ and $(1_W,1_Y)$, where I've used subscripts to indicate which of the three sets the element comes from. The equivalence relation thus consists of these two pairs, the pair $(1_X,1_Y)$, the three pairs obtained from these by reversing the order of components, and all pairs of the form $(a,a)$ for $a\in X\sqcup Y\sqcup W$. When you form the quotient by this equivalence relation, you identify the $1$'s of the three given sets, but there are no further identifications. In particular, distinct elements in the $[0..1)$ in $X$ and the $(1..2]$ in $Y$ remain distinct in the pushout.

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I see. My mistake was in calculating the equivalence relation. Although the equivalence relation is a subset of the cartesian product, the reflexive, symmetric, and transitive properties must be tested against every element of the host set. – Reb.Cabin Jul 31 '13 at 16:54

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