Bicartesian squares (both pull-back and push-out) and Short exact sequences

Bicartesian Squares and Short Exact Sequences in an Abelian CategoryPropositionPullback as kernelThe cokernel half by dualityRemark: Why a pullback is an equalizer over a product

 

Bicartesian Squares and Short Exact Sequences in an Abelian Category

Let $\mathcal{A}$$\mathcal A$ be an abelian category. Consider a commutative square

$$\begin{array}{ccc}P& \stackrel{i}{\to }& X\\ \downarrow j& & \downarrow u\\ Y& \stackrel{v}{\to }& Z.\end{array}$$

Thus

$$ui=vj.$$

Since $\mathcal{A}$$\mathcal A$ is additive, finite products and finite coproducts coincide. Hence $X\oplus Y$$X\oplus Y$ is a biproduct. Define

$$s=(\genfrac{}{}{0ex}{}{i}{-j}):P\to X\oplus Y,$$

and

$$d=[uv]:X\oplus Y\to Z.$$

By commutativity of the square,

$$ds=[uv](\genfrac{}{}{0ex}{}{i}{-j})=ui-vj=0.$$

So every commutative square canonically gives a complex

$$P\stackrel{(\genfrac{}{}{0ex}{}{i}{-j})}{\to }X\oplus Y\stackrel{[uv]}{\to }Z.$$

The main point is that this complex is short exact precisely when the original square is bicartesian.

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Proposition

Let

$$\begin{array}{ccc}P& \stackrel{i}{\to }& X\\ \downarrow j& & \downarrow u\\ Y& \stackrel{v}{\to }& Z\end{array}$$

be a commutative square in an abelian category. Then the square is bicartesian if and only if the associated sequence

$$0⟶P\stackrel{(\genfrac{}{}{0ex}{}{i}{-j})}{\to }X\oplus Y\stackrel{[uv]}{\to }Z⟶0$$

is short exact.

Equivalently,

$$\begin{array}{ccc}P& \to & X\\ \downarrow & & \downarrow \\ Y& \to & Z\end{array}\text{ is bicartesian}⟺(\genfrac{}{}{0ex}{}{i}{-j})=\ker [uv],[uv]=\mathrm{coker}(\genfrac{}{}{0ex}{}{i}{-j}).$$

We prove the kernel half. The cokernel half follows by the duality principle.

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Pullback as kernel

We first prove:

$$\boxed{{\displaystyle \text{The square is a pullback}⟺(\genfrac{}{}{0ex}{}{i}{-j}):P\to X\oplus Y\text{ is a kernel of }[uv]:X\oplus Y\to Z.}}$$

The conceptual reason is simple:

$$\text{pullback}=\text{equalizer over a product},$$

and in an additive category,

$$\text{equalizer}=\text{kernel of a difference}.$$

Let us spell this out.

The pullback of the cospan

$$X\stackrel{u}{\to }Z\stackrel{v}{\leftarrow }Y$$

can be described as the equalizer of the two morphisms

$$u{\pi }_X,v{\pi }_Y:X\times Y\to Z,$$

where

$${\pi }_X:X\times Y\to X,{\pi }_Y:X\times Y\to Y$$

are the product projections.

Since we are in an abelian category,

$$X\times Y\cong X\oplus Y.$$

Under this identification, the two maps become

$$u{\pi }_X,v{\pi }_Y:X\oplus Y\to Z.$$

In an additive category, the equalizer of two morphisms $f,g:A\to B$$f,g:A\to B$ is the kernel of their difference:

$$\mathrm{Eq}(f,g)=\ker (f-g).$$

Therefore the pullback of

$$X\stackrel{u}{\to }Z\stackrel{v}{\leftarrow }Y$$

is the kernel of

$$u{\pi }_X-v{\pi }_Y:X\oplus Y\to Z.$$

But

$$u{\pi }_X-v{\pi }_Y=[u-v].$$

Hence the pullback is

$$\ker [u-v].$$

Now the original square gives a morphism

$$(\genfrac{}{}{0ex}{}{i}{j}):P\to X\oplus Y.$$

Indeed,

$$[u-v](\genfrac{}{}{0ex}{}{i}{j})=ui-vj=0.$$

Thus the original square is a pullback precisely when

$$(\genfrac{}{}{0ex}{}{i}{j}):P\to X\oplus Y$$

is a kernel of

$$[u-v]:X\oplus Y\to Z.$$

This is almost the desired statement. The only difference is the sign convention. We want the map

$$(\genfrac{}{}{0ex}{}{i}{-j}):P\to X\oplus Y$$

and the map

$$[uv]:X\oplus Y\to Z.$$

Let

$$\tau ={\mathrm{id}}_X\oplus (-{\mathrm{id}}_Y):X\oplus Y\to X\oplus Y.$$

This is an isomorphism. Moreover,

$$\tau (\genfrac{}{}{0ex}{}{i}{j})=(\genfrac{}{}{0ex}{}{i}{-j}),$$

and

$$[uv]\tau =[u-v].$$

Therefore,

$$(\genfrac{}{}{0ex}{}{i}{j})=\ker [u-v]⟺(\genfrac{}{}{0ex}{}{i}{-j})=\ker [uv].$$

Hence the square is a pullback if and only if

$$(\genfrac{}{}{0ex}{}{i}{-j}):P\to X\oplus Y$$

is a kernel of

$$[uv]:X\oplus Y\to Z.$$

This proves the kernel half.

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The cokernel half by duality

Now apply the duality principle.

Passing to the opposite category reverses all arrows. Under this operation:

• pullback $\leftrightarrow$$\leftrightarrow$ pushout,

• kernel $\leftrightarrow$$\leftrightarrow$ cokernel,

and biproducts remain biproducts.

Therefore the dual of the previous statement is:

$$\boxed{{\displaystyle \text{The square is a pushout}⟺[uv]:X\oplus Y\to Z\text{ is a cokernel of }(\genfrac{}{}{0ex}{}{i}{-j}):P\to X\oplus Y.}}$$

So we obtain both halves:

$$\text{pullback}⟺(\genfrac{}{}{0ex}{}{i}{-j})=\ker [uv],$$

and

$$\text{pushout}⟺[uv]=\mathrm{coker}(\genfrac{}{}{0ex}{}{i}{-j}).$$

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Remark.

$$\begin{array}{ccc}P& \stackrel{i}{\to }& X\\ \downarrow & & \downarrow u\\ 0& ⟶& Z.\end{array}⟺0\to P\stackrel{i}{\to }X\stackrel{u}{\to }Z\to 0\text{ exact}$$

Remark: Why a pullback is an equalizer over a product

Let $\mathcal{C}$$\mathcal C$ be a category with binary products and equalizers. Consider a cospan

$$X\stackrel{u}{\to }Z\stackrel{v}{\leftarrow }Y.$$

Its pullback is usually denoted by

$$X{\times }_{Z}Y.$$

The key observation is that a point of the pullback should be thought of as a pair consisting of one piece of data in $X$$X$ and one piece of data in $Y$$Y$, subject to the compatibility condition that their images in $Z$$Z$ agree.

Categorically, the object that first collects arbitrary pairs of data from $X$$X$ and $Y$$Y$ is the product

$$X\times Y.$$

Let

$${\pi }_X:X\times Y\to X,{\pi }_Y:X\times Y\to Y$$

be the product projections. From these we get two morphisms

$$u{\pi }_X,v{\pi }_Y:X\times Y\to Z.$$

The first sends a pair to its $X$$X$-component and then into $Z$$Z$; the second sends the same pair to its $Y$$Y$-component and then into $Z$$Z$. The compatible pairs are precisely those for which these two morphisms agree.

Therefore the pullback should be the equalizer

$$\mathrm{Eq}(u{\pi }_X,v{\pi }_Y)⟶X\times Y.$$

Let us verify this by the universal property.

Suppose

$$E\stackrel{e}{\to }X\times Y$$

is the equalizer of

$$u{\pi }_X,v{\pi }_Y:X\times Y\to Z.$$

Define

$$p_X={\pi }_{X}e:E\to X,p_Y={\pi }_{Y}e:E\to Y.$$

Since $e$$e$ equalizes $u{\pi }_X$$u\pi_X$ and $v{\pi }_Y$$v\pi_Y$, we have

$$u{p}_X=u{\pi }_{X}e=v{\pi }_{Y}e=v{p}_Y.$$

Thus $E$$E$ gives a cone over the cospan

$$X\stackrel{u}{\to }Z\stackrel{v}{\leftarrow }Y.$$

Now let $T$$T$ be any object with morphisms

$$x:T\to X,y:T\to Y$$

such that

$$ux=vy.$$

By the product universal property, the pair $(x,y)$$(x,y)$ determines a unique morphism

$$⟨x,y⟩:T\to X\times Y$$

such that

$${\pi }_X⟨x,y⟩=x,{\pi }_Y⟨x,y⟩=y.$$

The compatibility condition $ux=vy$$ux = vy$ says exactly that

$$u{\pi }_X⟨x,y⟩=v{\pi }_Y⟨x,y⟩.$$

Hence the morphism $⟨x,y⟩:T\to X\times Y$$\langle x, y \rangle : T \to X \times Y$ equalizes $u{\pi }_X$$u\pi_X$ and $v{\pi }_Y$$v\pi_Y$. By the equalizer universal property, there exists a unique morphism

$$r:T\to E$$

such that

$$er=⟨x,y⟩.$$

Composing with the product projections gives

$$p_{X}r={\pi }_{X}er={\pi }_X⟨x,y⟩=x,$$

and

$$p_{Y}r={\pi }_{Y}er={\pi }_Y⟨x,y⟩=y.$$

Thus every compatible pair of maps $T\to X$$T \to X$ and $T\to Y$$T \to Y$ factors uniquely through $E$$E$. This is exactly the pullback universal property.

Therefore

$$X{\times }_{Z}Y\cong \mathrm{Eq}(u{\pi }_X,v{\pi }_Y).$$

In words:

$$\boxed{{\displaystyle \text{pullback}=\text{compatible pairs}=\text{equalizer inside the product}.}}$$

In an additive category, equalizers can be written as kernels of differences. Hence, if the category is additive, this becomes

$$X{\times }_{Z}Y\cong \ker (u{\pi }_X-v{\pi }_Y).$$

This is the step used above when we identify the pullback with a kernel.