Introduction to limits and colimits

Terminal objects and products(17)
Terminal object(1)

terminal object in a category \(\mathcal{C}\)

  • An object z is terminal if, for each object c there exists a unique morphism \(c \xrightarrow{!} z\)

  • We say terminal objects have a universal property

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Product(1)

Given two objects \(X,Y \in \mathcal{C}\), the product \(X \times Y\)

  • This is another object in \(\mathcal{C}\) with morphisms \(X \xleftarrow{p_x}X\times Y\xrightarrow{p_y}Y\)

  • This should satisfy the property that there exists a unique morphism making the following diagram commute for any other object C and morphisms \(X \xleftarrow{f}C\xrightarrow{g}Y\)

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Terminal in Set(1)

In Set, any set with exactly one element is a terminal object. For an arbitrary other set, we have no choice but to send everything to that one object when specifying a function.

Product in Set(1)

  • In Set, the categorical product of two sets is our usual cartesian product.

  • The projections are \(x \xrightarrow{p_x}(x,y)\xrightarrow{p_y}y\)

  • The unique morphism from some \(X \xleftarrow{f} C \xrightarrow{g} Y\), the unique map \(C \xrightarrow{!}X \times Y\) is given by \((f\times g)(c):=(f(c),g(c))\).

Product in Cat(1)

The product of two categories \(\mathcal{C}\times \mathcal{D}\) may be given as follows:

  • \(Ob(C\times D)\) are the pairs \((c,d)\) where c is an object of \(\mathcal{C}\) and d is an object of \(\mathcal{D}\).

  • Morphisms are pairs \((c,d)\xrightarrow{(f,g)}(c',d')\) where \(c \xrightarrow{f}c'\) is a morphism in \(\mathcal{C}\) and \(d \xrightarrow{g}d'\) is a morphism in \(\mathcal{D}\).

  • Composition is given by composing each entry in the pair separately.

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Terminal objects are isomorphic(2)
Proof(1)

  • Suppose \(Z,Z'\) are both terminal objects. Therefore there exist unique maps \(Z \overset{a}{\underset{b}{\rightleftarrows}}Z'\)

  • Composing these we get \(Z \xrightarrow{a;b} Z\), but this is forced to be the identity map because there is only one morphism from \(Z\) to itself and we have to have an identity.

  • Therefore we can talk about ’the terminal object’ as if there were only one.

All terminal objects in a category \(\mathcal{C}\) are isomorphic

Exercise 3-81(2)

Let \(z \in P\) be an element of a preorder, and consider the corresponding category \(\mathcal{P}\). Show that z is a terminal object iff it is a top element in \(P\), i.e. \(\forall c \in P: c \leq z\)

Solution(1)

There is a morphism from every object if every object is less than z, and the uniqueness comes from the fact that preorders are thin categories.

Exercise 3-82(2)

Name a terminal object in the category Cat

Solution(1)

1 is terminal because a functor from any other category is forced to map all objects to 1 and all morphisms to its identity morphism.

Exercise 3-83(2)

Name a category which does not have a terminal object

Solution(1)

The category corresponding to the natural numbers has no terminal object (it would be an integer larger than all integers).

Exercise 3-88(2)

Let \(x,y \in P\) be elements of a preorder and \(\mathcal{P}\) be the corresponding category. Show that the product \(x \times y\) in \(\mathcal{P}\) agrees with their meet \(x \land y\) in \(P\).

Solution(1)

  • The uniqueness aspect of the product is not relevant since all morphisms are unique in a preorder category.

  • The product definition translates to an operation which takes a pair of objects in a preorder and gives us another object with the property that \(x \times y \leq x\) and \(x \times y \leq y\), and for any other b that also has this property we have \(b \leq x\times y\)

  • Considering the set \(A=\{x,y\}\), the conditions for \(x \times y\) matches the definition of \(\bigwedge A\) (grestest lower bound).

Exercise 3-90(2)

  1. What are identity morphism in a product category \(\mathcal{C}\times \mathcal{D}\)?

  2. Why is composition in a product category associative?

  3. What is the product category \(1 \times 2\)?

  4. What is the product category of two preorders?

Solution(1)

  1. For object \((c,d)\), the identity morphism is \((id_c,id_d)\)

  2. The operation was defined in terms of function composition which is associative.

  3. It is isomorphic to just 2

  4. The underlying set is the cartesian product, and \((a,b)\leq(a',b')\) iff \(a \leq a' \land b \leq b'\)

Limits(8)
Free diagram(1)

  • Suppose \(\mathcal{J}\) is the free category on the graph \(\boxed{1 \rightarrow 2 \leftarrow 3 \rightrightarrows 4 \rightarrow 5}\)

  • We may draw a diagram \(\mathcal{J}\xrightarrow{D}\mathcal{C}\) inside \(\mathcal{C}\) as below:

  • \(\boxed{D_1 \rightarrow D_2 \leftarrow D_3 \rightrightarrows D_4 \rightarrow D_5}\)

  • We can represent this diagram as a cone over the diagram by picking a \(C \in \mathcal{C}\) for which every pair of parallel paths that start from \(C\) are the same.

Terminal object as limit(1)

Terminal objects are imits where the indexing category is empty, \(\mathcal{J}=0\).

Product as limit(1)

Products are limits where the indexing category consists of two objects with no arrows, \(\mathcal{J}=\boxed{\overset{v}\bullet\ \overset{w}\bullet}\).

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Cone(1)

A cone \((C,c_*)\) over a diagram \(\mathcal{J}\xrightarrow{D}\mathcal{C}\) and the category \(\mathbf{Cone}(D)\)

  • We require:

    • An object \(C \in Ob(\mathcal{C})\)

    • For each object \(j \in \mathcal{J}\), a morphism \(C \xrightarrow{c_j}D(j)\).

  • The following property must be satisfied:

    • \(\forall f \in \mathcal{J}(j,k):\) \(c_k=c_j;D(f)\)

  • A morphism of cones is a morphism \(C \xrightarrow{a} C'\) in \(\mathcal{C}\) such that, for all \(j \in \mathcal{J}\), we have \(c_j=a;c'_j\).

  • Cones and their morphisms form a category.

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Limit(1)

The limit of a diagram \(D\), \(lim(D)\)

  • The limit of \(D\), denoted is the terminal object in the category \(\mathbf{Cone}(D)\)

  • If \(lim(D)=(C,c_*)\) we refer to \(C\) as the limit object and the map \(c_j\) as the jth projection map

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Exercise 3-91(2)

Check that the product \(X \xleftarrow{p_X} X \times Y \xrightarrow{p_Y} Y\) is terminal object in \(\mathbf{Cone}(X,Y)\)

Solution(1)

The existence and uniqueness of the morphism to the product object in the product definition are the conditions of being a terminal object in \(\mathbf{Cone}(X,Y)\). The maps of the terminal object to \(X\) and \(Y\) are the projection maps of the product.

Finite limits in Set(9)
Limit formula(2)

  • Let \(\mathcal{J}\) be a category presented by the finite graph \((\{v_1,...,v_n\},A,s,t)\) with some equations.

  • Let \(\mathcal{J}\xrightarrow{D}\mathbf{Set}\) be some set-valued functor.

  • The set \(\underset{\mathcal{J}}{lim}D := \{(d_1,...,d_n)\ |\ \forall i: d_i \in D(v_i)\ \text{and}\ \forall v_i\xrightarrow{a}v_j \in A: D(a)(d_i)=d_j\}\)

    • ... together with projection maps \(lim_\mathcal{J}D \xrightarrow{p_i}D(v_i)\) given by \(p_i(d_1,...,d_n):=d_i\)

  • ... is a limit of \(D\). NOCARD

Proof(1)

NOT PROVEN

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Terminal object in limit formula(1)

  • With respect to Proposition 3.95, if \(\mathcal{J}=0\), then \(n=0\); there are no vertices.

  • There is exactly one empty tuple which vacuously satisfies the properties, so we’ve constructed the limit as the singleton set \(\{()\}\) consisting of just the empty tuple.

  • Thus the limit of the empty diagram, i.e. the terminal object in Set, is the singleton set.

Pullback diagram(1)

  • If \(\mathcal{J}\) is presented by the cospan graph \(\boxed{\overset{x}\bullet \xrightarrow{f} \overset{a}\bullet \xleftarrow{g}\overset{y}\bullet}\) then its limit is known as a pullback.

  • Given the diagram \(X \xrightarrow{f}A\xleftarrow{g}Y\), the pullback is the cone shown below:

  • Because the diagram commutes, the diagonal arrow is superfluous. One can denote pullbacks instead like so:

  • The pullback picks out the \((X,Y)\) pairs which map to the same output.

Exercise 3-97(2)

  • Show that the limit formula works for products in Set

  • The diagram, whose limit is a product, is \(\mathcal{J}=\boxed{\overset{v}\bullet\ \overset{w}\bullet}\) (see Exaample 3.94)

Solution(1)

  • \(V=\{v,w\}, A=\varnothing\)

  • The second condition of the set comprehension is vacuously satisfied because \(A = \varnothing\)

  • So all we have left is all pairs of tuples where the first element comes from \(D(v)\) and the second element comes from the set \(D(w)\).

  • This is the Cartesian product \(D(v) \times D(w)\)

Exercise 3-99(2)

If \(1 \xrightarrow{D}\mathbf{Set}\) is a functor, what is the limit of \(D\)? Compute using the limit formula and check answer against the limit definition.

Solution(1)

  • There are no arrows, so we just recover the set \(D(1)\) as the limit.

  • The limit definition first requires the category \(\mathbf{Cone}(1)\)

    • There is only one possible cone, so \(Cone(1)\cong 1\)

  • The terminal object in \(1\) is the sole object of the category.

A brief note on colimits(3)
Cocone(1)

A cocone in a category \(\mathcal{C}\)

  • A cone in \(\mathcal{C}^{op}\)

  • Given a diagram \(\mathcal{J}\xrightarrow{D}\mathcal{C}\), we may take the limit of the functor \(\mathcal{J}^{op}\xrightarrow{D^{op}}\mathcal{C}^{op}\) is a cocone in \(\mathcal{C}\) - the colimit of \(D\) is this cocone.

Exercise 3-101(2)

Let \(\mathcal{C}\xrightarrow{F}\mathcal{D}\) be a functor. How should we define its opposite: \(\mathcal{C}^{op}\xrightarrow{F^{op}}\mathcal{D}^{op}\)

Solution(1)

  • There is an isomorphism between a category and its opposite, meaning there are natural functors \(\overset{\cong}\rightarrow\) which alternate between them.

  • Define \(\mathcal{C}^{op}\xrightarrow{F^{op}}\mathcal{D}^{op}\) as \(F ; \overset{\cong}\rightarrow\). This is a valid functor as it is the composition of two functors.