A detailed proof required.

Question

The question is given below:

Prove that in $S \bigsqcup T,$ it must be that $S \bigcup T = S \bigsqcup T$

Hints I received are:

1-To write the question as: in $S \bigsqcup T,$ it must be that $ S\sqcup T=\textrm{in}_S(S)\cup \textrm{in}_T(T). $\

2- Take $U = S \bigcup T, f = in_S $ and $g = in_T.$

3- the problem will be in the uniqueness of $(f,g),$ you can find a function that breaks this uniqueness.

My question is:

I do not understand how are hints 2 and 3 can be used to create a full solution, could anyone help me in this please?

Answer 1

Suppose we have ( in the category $\textsf{Set}$) some set $S \coprod T$ and maps $\text{in}_S$ and $\text{in}_T$ that obey the universal property for coproduct (sum).

First we can see that the $\text{in}_S$ map is injective. Suppose not and we have $s_1\neq s_2$ in $S$ with $\text{in}_S(s_1) = \text{in}_S(s_2)=p \in S \coprod T$. Then define $U=\{0,1\}$ and $f$ is defined on $S$ as being $0$ on $s_1$ and $1$ for all other points. $ g: T \to U$ is just constantly $0$, it’s not important. Then there should be a map $ f \coprod g$ completing the diagram uniquely. But

$$(f \coprod g)(p)= ((f \coprod g) \circ \text{in}_S)(s_1) = f(s_1)=0$$ and

$$(f \coprod g)(p) = ((f \coprod g) \circ \text{in}_S)(s_2)= f(s_2)=1$$

And this is a contradiction. In a completely analogous way we can show that $\text{in}_T$ is injective.

Then we can show that $$\text{in}_S[S] \cap \text{in}_T[T] =\emptyset$$

which partly justifies the name disjoint union for the coproduct. Suppose some $p$ lies in the intersection. So we can write $p = \text{in}_S(s_0) = \text{in}_T(t_0)$ for some $s_0\ in S, t_0\in T$.

We again use $U=\{0,1\}$ and define $f: S \to U$ by $f(s_0)=0$ and $1$ elsewhere, and $g: T \to U$ by g(t_0)=1$ and $0$ elsewhere. If $f \coprod g$ exists completing the diagram, then similarly:

$$(f \coprod g)(p)= (f \coprod g) \circ \text{in}_S)(s_0) = f(s_0) = 0$$

and also

$$(f \coprod g)(p)= (f \coprod g) \circ \text{in}_T)(t_0) = g(t_0) = 1$$

which is a contradiction and $f\coprod g $ cannot exist, again.

Now the claim is that $$\text{in}_S[S] \cup \text{in}_T[T] = S \coprod T$$

and again we assume this is not the case, and we have some $p$ in $S \coprod T$ that is not in the left hand union of subsets. Let $h: S \coprod T \to S \coprod T$ be defined by $h(p)=q$ for some $S \coprod T\ni q \neq p$ and $h(x)=x$ otherwise. Then both $1_{S \coprod T}$ and $h$ (and these are different maps!) make that the diagram for $f= \text{in}_S$ and $g = \text{in}_T$ commutes. So in that case the map from the universal property is not unique, which cannot happen. So $S \coprod T$ is a disjoint union of the injective images $\text{in}_S[S]$ and $\text{in}_T[T]$ of $S$ resp. $T$.

As usual with these objects defined by universal properties, the object is uniquely determined up to isomorphism (bijection in this category) and the standard construction $S \times \{0\} \cup T \cup \{1\}$ with $\text{in}_S(s)=(s,0)$ and $\text{in}_T(t)=(t,1)$ is one way to create an instance of this coproduct. But the universality itself already implies many of its properties, if we pick the right objects and maps to apply it to. A mixture of "point-based" and category theory based arguments is needed here.

Answer 2

It seems to me that you must prove that $S\cup T$ does not work in general as coproduct of $S$ and $T$.

This because things can go wrong if $S\cap T\neq\varnothing$.

If $S\cap T\neq\varnothing$ then some $x\in S\cap T$ might exists with $f(x)\neq g(x)$.

This avoids the existence of a function $h:S\cup T\to U$ that satisfies $h\circ\mathsf{in}_S=f$ and $h\circ\mathsf{in}_T=g$.

This because:

$$(f\circ\mathsf{in}_S)(x)=f(x)\neq g(x)=(g\circ\mathsf{in}_T)(x)$$

Answer 3

You might want to be a bit careful about the union when working with universal properties. Indeed object defined by universal properties are always unique up to isomorphism, and the union is not at all invariant under isomorphism.

Let me explain : If you consider $S = \left\{ a,b,c \right\}$, $T = \left\{ a,x \right\}$ and $T'=\left\{ y,z \right\}$, then $T$ and $T'$ are isomorphic (i.e. in bijection), so necessarily $S\sqcup T$ and $S\sqcup T'$ are isomorphic. However direct calculation shows $S\cup T= \left\{ a,b,c,x \right\}$ and $S\cup T' = \left\{ a,b,c,y,z \right\}$. So you can immediately see that $S\cup T$ and $S\cup T'$ are not isomorphic.

So this shows that the property you want to show cannot hold, as $S\cup T$ can never be defined by a universal property. However, the hints you have given are correct, in the sense that taking $U = S\cup T$, you can notice that you have two maps $S\to S\cup T$ and $T\to S\cup T$, which lets you build a map $S\sqcup T\to S\cup T$. But this map is not an isomorphism. In fact you can construct $S\sqcup T$ explictly as the disjoint union of $S$ and $T$ and from there you can see that the map $S\sqcup T\to S\cup T$ is always surjective, but might not be injective when $S$ and $T$ are not disjoint.

As an additional note: The correct way to handle usual union categorically is to formulate it without referring to the name of the elements, and build it as a pushout of the two monomorphisms $S\cap T \hookrightarrow S$ and $S\cap T\hookrightarrow T$.