Limits and Colimits: Examples and Counterexamples

Coordinate evaluation gives functions $p_i:{\prod }_i{A}_i\to A_i$. Given $f_i:X\to A_i$, define $f(x)=(f_i(x){)}_i$. Then $p_{i}f=f_i$, and these equations determine every value of $f$, proving existence and uniqueness in [F1].

If $I=\mathrm{\varnothing }$, the product consists of the single empty function, so there is exactly one function from every $X$ to it. Thus the same verification covers the nullary product.

Let the $i$th injection send $x$ to $(i,x)$. Given $g_i:A_i\to X$, define $g(i,x)=g_i(x)$. This is the unique function whose composite with the $i$th injection is $g_i$ for all $i$, so [F1] proves the coproduct property. For $I=\mathrm{\varnothing }$, the tagged union is empty and has exactly one function to every set.

The inclusion $e:E\hookrightarrow X$ satisfies $fe=ge$. If $h:Z\to X$ equalizes $f,g$, every $h(z)$ lies in $E$, so $h$ has a unique corestriction $Z\to E$. This is the equalizer property [F1].

Intersect all equivalence relations on $Y$ containing the pairs $(f(x),g(x))$; by [F3] the result $\sim$ is the least one. Its quotient map $q$ satisfies $qf=qg$.

If $h:Y\to Z$ satisfies $hf=hg$, equality of $h$ is itself preserved under reflexive, symmetric, and transitive closure, so $h$ is constant on $\sim$-classes. By [L1] it factors uniquely through $q$. Conversely every map through $q$ equalizes $f,g$. Thus $q$ is the coequalizer in [F1].

The coordinate projections of the displayed subset satisfy the cospan equation. If $r:T\to X$ and $s:T\to Y$ satisfy $fr=gs$, then $t\mapsto (r(t),s(t))$ is the unique function into the subset with those two projections. This is [F1].

In the tagged union, generate an equivalence relation from $(0,a(w))\sim (1,b(w))$. The quotient injections agree on $W$.

Compatible maps $r:X\to T$ and $s:Y\to T$ define a function on the tagged union by cases. It is equal on every generating pair and hence constant on classes, so [F3] gives a unique quotient factor. Conversely any quotient map restricts to such a compatible pair. This is the pushout property [F1].

The restricted coordinate projections $P\to X,Y$ are continuous by [F2] and [F3], and their composites with $f,g$ agree by definition of $P$.

If $r:T\to X$ and $s:T\to Y$ are continuous with $fr=gs$, their unique Set-theoretic pairing lands in $P$ by [L1]. Its composite with $P\hookrightarrow X\times Y$ is $(r,s)$, continuous by [F2], so [F3] makes the factor $T\to P$ continuous.

Its uniqueness follows from uniqueness of its two coordinate functions. Thus [F1] identifies the displayed space as the pullback.

Since $f(1)=g(1)=1$, the identity lies in $E$. If $x,y\in E$, then $f(xy)=f(x)f(y)=g(x)g(y)=g(xy)$, and similarly $f(x^{-1})=g(x^{-1})$. Thus $E$ is a subgroup and its inclusion is a homomorphism.

The inclusion equalizes $f,g$. If $h:K\to G$ satisfies $fh=gh$, then $h(K)\subseteq E$, so the unique set-theoretic corestriction $\bar{h}:K\to E$ is a homomorphism and the inclusion composed with $\bar{h}$ is $h$.

Injectivity of the inclusion makes this factor unique. By [F1], it is the equalizer.

The inclusions $i_n:X_n\to X$ form a cocone. Let $f_n:X_n\to Y$ be any cocone, so $f_m{\mathrm{\mid }}_{X_n}=f_n$ whenever $n\le m$.

For $x\in X$, choose any $n$ with $x\in X_n$ and set $f(x)=f_n(x)$. If $x\in X_n\cap X_m$, then at the later stage $r=\max \{n,m\}$ the cocone equations give $f_n(x)=f_r(x)=f_m(x)$. Thus $f$ is well-defined.

The equations $f{i}_n=f_n$ hold by definition and determine $f$ on every element of the union, so the factor is unique. By [F1], $X$ is the colimit.

In [L1], tagged copies of one element appearing at different stages are identified at a common later stage. Hence its quotient is canonically the same ordinary union, without any assumption that the stages are disjoint.

A cone from $x$ to $(p_i)$ is exactly the assertion $x\le p_i$ for all $i$. Its unique factor through $p$ says $x\le p$ for every lower bound $x$. Thus [F2] is exactly the greatest-lower-bound property.

Reversing all inequalities turns the coproduct property into the least-upper-bound property. The empty cases give the greatest and least elements, respectively.

If $f,g:x\rightrightarrows y$ exist, [F1] gives $f=g$. Then $1_x$ equalizes them and every arrow into $x$ factors through $1_x$ uniquely. Dually, $1_y$ is a coequalizer.

For every set $X$, exactly one function $X\to \{\ast \}$ and exactly one function $\mathrm{\varnothing }\to X$ exist. Thus the singleton is terminal and the empty set initial in $\mathbf{S}\mathbf{e}\mathbf{t}$.

For every group $G$, the constant map $G\to 1$ is the unique homomorphism to the trivial group. A homomorphism $1\to G$ must send the identity to the identity, so it too is unique. Thus $1$ is both terminal and initial in $\mathbf{G}\mathbf{r}\mathbf{p}$.

Applying [L1] to steps 1.1 and 1.2 gives the four claimed empty-diagram limits and colimits.

The ordinary Cartesian product of any set-indexed family of nonempty sets is nonempty by [F3] and has the product property [F1] inside the full subcategory. For the empty family, the singleton is a nonempty terminal object. Thus ${\mathbf{S}\mathbf{e}\mathbf{t}}_{\mathrm{\ne }\mathrm{\varnothing }}$ has all small products.

Let $f,g:\{\ast \}\rightrightarrows \{0,1\}$ be the constant maps with values $0$ and $1$. If $h:X\to \{\ast \}$ equalized them, then $fh$ and $gh$ would be the distinct constant functions on the nonempty set $X$. Hence no equalizing cone exists in ${\mathbf{S}\mathbf{e}\mathbf{t}}_{\mathrm{\ne }\mathrm{\varnothing }}$, so in particular no equalizer [F2] exists.

The parallel-pair diagram is finite and small but has no limit, refuting the statement. This is exactly the missing equalizer data isolated by [L1].

Define the monotone map $F:P\to Q$ by $F(0)=0$ and $F(a)=F(b)=F(1)=1$. By [F3] it is a functor. Every arrow in either poset category is monic, since two parallel arrows are automatically equal; hence $F$ preserves every monomorphism.

In $P$, the pullback of $a\to 1\leftarrow b$ is the meet $a\wedge b=0$, as follows directly from [F1] after translating arrows to inequalities. Its image is $0$.

The image cospan in $Q$ is $1\to 1\leftarrow 1$, whose pullback is $1$. The canonical comparison is the noninvertible arrow $0\to 1$, so [L1] says that $F$ does not preserve this pullback. This refutes the statement.

The inclusion $i:Q\hookrightarrow P$ is full: for retained objects, an arrow exists in $Q$ exactly when it exists in $P$.

In $P$, the greatest common lower bound of $a,b$ is $m$, so [F1] makes $m=a\times b$. In $Q$, the object $m$ is absent and the greatest common lower bound of $a,b$ is $q$, so $q$ is their product there.

The image under $i$ of the product cone in $Q$ has apex $q$, whereas the ambient product has apex $m$. Its canonical comparison is $q\to m$, not an isomorphism because $m\nleqq q$. By [L1], the full inclusion does not preserve this product, refuting the statement.

The positive-integer chain is nonempty, two indices have their maximum as a common target, and it has no distinct parallel arrows, so it is filtered by [F1]. For fixed $i$, the countable product ${\prod }_{j\ge 1}{M}_{i,j}$ is the set of positive-integer sequences all of whose entries are at most $i$.

For each $j$, the filtered colimit ${\mathrm{colim}}_i{M}_{i,j}$ is ${\mathbb{N}}_{>0}$. The product of these coordinatewise colimits is the set of all positive-integer sequences, including $(1,2,3,\dots )$, which is unbounded.

Its filtered colimit over $i$ is therefore the set of bounded positive-integer sequences: a sequence appears at some stage exactly when one integer bounds all its coordinates.

Hence the canonical map from step 2.1 to the set in step 1.2 is not surjective. The arbitrary-product assertion is false, while [L1] is not contradicted because the product is infinite.

Let $f,g:\{\ast \}\rightrightarrows \{0,1\}$ be constant at $0$ and $1$. For any nonempty set $X$, the unique function $X\to \{\ast \}$ has composites constant at different values, so it does not equalize $f,g$.

Thus this parallel-pair diagram has no cone and in particular no equalizer [F2]. Its indexing category is finite and hence small.

By [F1], the category of nonempty sets is not complete. This category refutes the universal statement.

The binary product of the sole object of $\mathbf{1}$ with itself is that object, and $\mathrm{\varnothing }\times \mathrm{\varnothing }=\mathrm{\varnothing }$. The image product cone consists of identity functions on $\mathrm{\varnothing }$, so $F$ preserves the binary product.

Every parallel pair in $\mathbf{1}$ is $(1,1)$ and has identity equalizer. Its image is $(1_{\mathrm{\varnothing }},1_{\mathrm{\varnothing }})$, whose identity is also an equalizer. Thus $F$ preserves equalizers.

The sole object of $\mathbf{1}$ is terminal, but $\mathrm{\varnothing }$ is not terminal in $\mathbf{S}\mathbf{e}\mathbf{t}$ because no function $\{\ast \}\to \mathrm{\varnothing }$ exists. So $F$ does not preserve the empty product, hence does not preserve all finite limits and is not continuous by [F1].

This refutes the statement and exhibits exactly the missing nullary case in [L1].

For each small Top-diagram, [L1] constructs its limit on exactly the underlying Set-limit and adds the initial topology. Applying $U$ removes that topology and leaves the Set-limit cone unchanged.

Therefore every image cone is limiting in $\mathbf{S}\mathbf{e}\mathbf{t}$, which is preservation by [F1]. The asserted counterexample cannot exist, so the statement is false.

This does not claim reflection: a Set-limiting underlying cone need not already carry the initial topology required for a Top-limit.

By [L1] and [L2], the empty-diagram colimit in $\mathbf{G}\mathbf{r}\mathbf{p}$ is its initial object, the trivial group. Its underlying set is a singleton.

The underlying diagram is still empty. Its Set-colimit is the initial set $\mathrm{\varnothing }$ by [L2], not a singleton.

Thus the underlying-set functor does not preserve even the empty colimit, and the universal statement is false.

There is one empty function $\mathrm{\varnothing }\to X$ for every set $X$, so these functions form a cone with apex $\mathrm{\varnothing }$.

Any cone with apex $Y$ includes a function from $Y$ to the empty-set factor. Such a function exists only when $Y=\mathrm{\varnothing }$. Hence every cone has empty apex.

Between any two empty apices there is exactly one function, and all leg equations hold automatically because the functions are empty. Therefore the cone of step 1.1 is terminal among cones and is a product by [F1].

The indexing family is a proper class, so [F2] classifies the diagram as large. Its having this limit is compatible with completeness being a claim only about all small diagrams.