What is the intuition behind pullbacks and pushouts? For example I know that for terminal objects kind of end a category, they are kind of last is some sense, and that a product is a kind of pair, but what about pullbacks and pushouts what are the reasoning behind this names?
2 Answers
Pullbacks generalise many common situations; they can be thought of as equationally defined sub-objects or as the subobjects of products that satisfy certain equations.
Here are a few examples of pullbacks off the top of my head:
- inverse images are pullbacks
- intersection of subsets is a pullback
- more generally, intersection of (copies of) structures embedded in common larger structure, is a pullback; e.g., see here
- equationally defined categories (including sets) are pullbacks
- E.g., the category of elements of a functor to sets is obtained via pullback
- E.g., the set of solutions to any equation in two unknowns, such as $3x + 2 = y$, is obtained by pullback
- Relations are essentially spans and then relation composition is given by pullback
- characteristic predicate for sets make certain squares pullback, and that condition is used to specify characteristic arrows and truth-objects in general categories
- (binary) pullbacks and a terminal object yield all (finite) limits
- E.g., products and equalisers are pullbacks
In contrast, whereas pullbacks let us intersect in a mutual context, pushouts let us union along a mutual shared context. E.g., the pushout of two graph homomorphisms $A \leftarrow I \rightarrow B$ is essentially the union $A \cup B$ but we 'identify' (glue, "make equal") all the pieces that orginate from $I$; i.e., we make the disjoint union but treat the images of $I$ as being "the same" and so do not repeat those parts.
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Pullbacks are fibred-products, i.e., a product with some compatibility restrictions. The terminology came from differential geometry when you really pull differential forms or their bundle on $B$ back to differential forms or their bundle on $A$ along immersion $A\to B$. Product $A\times B$ is just a special case when you pullback $$ \require{AMScd} \begin{CD} @. B\\ @. @V{!}VV\\ A@>{!}>> 1 \end{CD} $$ which the terminal object $1$ doesn't impose any restrictions, and get $$ \begin{CD} A\times B@>{\operatorname{proj}_2}>> B\\ @V{\operatorname{proj}_1}VV @V{!}VV\\ A@>{!}>> 1 \end{CD} $$
Dually we have pushouts as a kind of sum, subject to some constraint. Indeed, in Sets we have the disjoint union $$ \begin{CD} \varnothing@>{!}>> B\\ @V{!}VV @V{i_2}VV\\ A@>{i_1}>> A\amalg B \end{CD} $$ as the pushout of $\varnothing\to A,B$, and we also have $$ \begin{CD} A\cap B@>>> B\\ @VVV @VVV\\ A@>>> A\cup B \end{CD}. $$ I don't think "pushout" was coined before the late-1940s when category theory came along, and merely chosen because it is clearly opposite to "pullback" (a similar word "pushforward" existed in other context but that name was not chosen).
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afteer months, the restriction now make sense to me, thank you so much – geckos Aug 20 '20 at 12:41
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What is an exclamation point function from the empty set?? – Alper Dec 10 '22 at 12:53
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@Alper "!" stands for "unique". So here, there is a unique map taking the null set to the desired codomain – Panini Jan 24 '23 at 17:29