Categorical Structures and Adjunctions in Pointed Sets (Set*)

Let us consider a functor $F$$F$ from $\mathrm{Set}$$\mathrm{Set}$ to ${\mathrm{Set}}_{\ast }$$\mathrm{Set}_*$.

For a set $X$$X$, we define $F(X):=X\coprod \{{\ast }_x\}$$F(X):=X\coprod\{*_x\}$. For a function $f:X\to Y$$f:X\to Y$, extend it uniquely at $\ast$$*$ by $F(f)({\ast }_x)={\ast }_y$$F(f)(*_x)=*_y$.

Proposition. This functor is the left adjoint of the forgetful functor $U:{\mathrm{Set}}_{\ast }\to \mathrm{Set}$$U:\mathrm{Set}_*\to\mathrm{Set}$ . i.e.

$${\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(FX,Y_{\ast })\cong {\mathrm{Hom}}_{\mathrm{Set}}(X,U{Y}_{\ast })$$

Proof. Obviously. $◻$$\square$

By the previous blog, we see $f$$f$ is mono iff $f$$f$ is injective.

Remark. Readers should compare this to the adjoint we mentioned before.

Category of sets with base point is a really good category, it has $0$$0$ object, which is $\{\ast \}$$\{*\}$, it is complete and cocomplete, since it is the comma category over $\mathrm{Set}$$\mathrm{Set}$. Hence we have equalizer and coequalizer, hence we could build kernel and cockerel, and do some interesting things.

In general, if $\mathcal{C}$$\mathcal C$ is a category with $0$$0$ object, then we define the $0$$0$ morphism as follows:

$$0_{X,Y}:X\to 0\to Y$$

If the equalizer exists, then we could define $\ker (f):=\mathrm{Eq}(f,0)$$\mathrm{ker}(f):=\mathrm{Eq}(f,0)$. Similarly for cokernel.

Now let us back to ${\mathrm{Set}}_{\ast }$$\mathrm{Set}_*$. Easy to see that $\ker (f)=f^{-1}(\ast )$$\mathrm{ker}(f)=f^{-1}(*)$ and $\mathrm{coker}(f)=Y/\mathrm{Im}(f)$$\mathrm{coker}(f)=Y/\mathrm{Im}(f)$, hence $\ker (\mathrm{coker})(f)=\mathrm{Im}(f)$$\mathrm{ker(coker)}(f)=\mathrm{Im}(f)$.

It is not like Abelian category, we do not have $\ker (\mathrm{coker})(f)=\mathrm{coker}(\ker )(f)$$\mathrm{ker(coker)}(f)=\mathrm{coker(ker)}(f)$.

But we still have, for any monomorphism $k$$k$, $k$$k$ is the kernel of $Y/\mathrm{Im}(k)$$Y/\mathrm{Im}(k)$. Also, for any $f:X\to Y$$f:X\to Y$, $0\circ f=0,f\circ 0=0$$0\circ f=0,f\circ 0=0$.

The coproduction in ${\mathrm{Set}}_{\ast }$$\mathrm{Set_*}$ is $X\vee Y.$$X\vee Y.$ The product is $(X\times Y,({\ast }_x,{\ast }_y))$$(X\times Y,(*_x,*_y))$.

This category is ecriched to itself, the base point of ${\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X,Y)$$\mathrm{Hom_{\mathrm{Set}_*}}(X,Y)$ is $0_{X,Y}$$0_{X,Y}$.

The tensor product is smash product, i.e. $X\wedge Y=X\times Y/X\vee Y$$X\wedge Y=X\times Y/X\vee Y$. We have tensor-hom adjoint, i.e.

$${\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X\wedge Y,Z)\cong {\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X,Z^Y)$$

Proof.

Let's clarify the notation:

• $X\wedge Y$$X\wedge Y$ is the smash product, defined as $X\times Y/X\vee Y$$X\times Y/X\vee Y$, where $X\vee Y$$X\vee Y$ is the wedge sum

• $Z^Y$$Z^Y$ is the internal hom object, corresponding to the set of basepoint-preserving maps from $Y$$Y$ to $Z$$Z$

• The basepoint of ${\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X,Y)$$\mathrm{Hom}_{\mathrm{Set}_*}(X,Y)$ is the constant map $0_{X,Y}$$0_{X,Y}$ sending everything to the basepoint

Structure of the Smash Product

First, note that $X\wedge Y=X\times Y/X\vee Y$$X\wedge Y = X\times Y/X\vee Y$, where $X\vee Y=(X\times \{{\ast }_Y\})\cup (\{{\ast }_X\}\times Y)$$X\vee Y = (X\times\{*_Y\}) \cup (\{*_X\}\times Y)$. This means that in the smash product, all points that contain any basepoint are identified to a single point, which becomes the basepoint of $X\wedge Y$$X\wedge Y$.

Constructing the Isomorphism

We will define a bijection: $\mathrm{\Phi }:{\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X\wedge Y,Z)\to {\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X,Z^Y)$$\Phi: \mathrm{Hom}_{\mathrm{Set}_*}(X\wedge Y,Z) \to \mathrm{Hom}_{\mathrm{Set}_*}(X,Z^Y)$

Forward Direction

For any $f\in {\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X\wedge Y,Z)$$f \in \mathrm{Hom}_{\mathrm{Set}_*}(X\wedge Y,Z)$, we define $\mathrm{\Phi }(f)=g_f$$\Phi(f) = g_f$ where:

• $g_f:X\to Z^Y$$g_f: X \to Z^Y$ is given by $g_f(x)(y)=f([x,y])$$g_f(x)(y) = f([x,y])$

• Here $[x,y]$$[x,y]$ represents the equivalence class in $X\wedge Y$$X\wedge Y$

We need to verify that $g_f$$g_f$ preserves basepoints:

• $g_f({\ast }_X)(y)=f([{\ast }_X,y])=f({\ast }_{X\wedge Y})={\ast }_Z$$g_f(*_X)(y) = f([*_X,y]) = f(*_{X\wedge Y}) = *_Z$ (since $[{\ast }_X,y]={\ast }_{X\wedge Y}$$[*_X,y] = *_{X\wedge Y}$)

• Therefore $g_f({\ast }_X)=0_{Y,Z}$$g_f(*_X) = 0_{Y,Z}$, the constant map to ${\ast }_Z$$*_Z$

Reverse Direction

Conversely, for any $g\in {\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X,Z^Y)$$g \in \mathrm{Hom}_{\mathrm{Set}_*}(X,Z^Y)$, we define $\mathrm{\Psi }(g)=f_g$$\Psi(g) = f_g$ where:

• $f_g:X\wedge Y\to Z$$f_g: X\wedge Y \to Z$ is given by $f_g([x,y])=g(x)(y)$$f_g([x,y]) = g(x)(y)$

We need to verify that $f_g$$f_g$ is well-defined on equivalence classes:

• If $(x,y)\ne (x^{\prime },y^{\prime }),[x,y]=[x^{\prime },y^{\prime }]$$(x,y)\ne(x',y'),[x,y] = [x',y']$, then either $x={\ast }_X$$x=*_X$ or $y={\ast }_Y$$y=*_Y$

• If $x={\ast }_X$$x=*_X$, then $g(x)=g({\ast }_X)=0_{Y,Z}$$g(x) = g(*_X) = 0_{Y,Z}$, so $g(x)(y)={\ast }_Z$$g(x)(y) = *_Z$

• If $y={\ast }_Y$$y=*_Y$, then $g(x)(y)=g(x)({\ast }_Y)={\ast }_Z$$g(x)(y) = g(x)(*_Y) = *_Z$ (since $g(x)$$g(x)$ preserves basepoints)

• Therefore, $f_g$$f_g$ is well-defined on equivalence classes

Verification of Bijection

We now verify that $\mathrm{\Phi }$$\Phi$ and $\mathrm{\Psi }$$\Psi$ are mutual inverses:

1. $(\mathrm{\Phi }\circ \mathrm{\Psi })(g)(x)(y)=\mathrm{\Phi }(f_g)(x)(y)=f_g([x,y])=g(x)(y)$$(\Phi \circ \Psi)(g)(x)(y) = \Phi(f_g)(x)(y) = f_g([x,y]) = g(x)(y)$, so $\mathrm{\Phi }\circ \mathrm{\Psi }=\mathrm{id}$$\Phi \circ \Psi =\mathrm{id}$

2. $(\mathrm{\Psi }\circ \mathrm{\Phi })(f)([x,y])=f_{\mathrm{\Phi }(f)}([x,y])=\mathrm{\Phi }(f)(x)(y)=f([x,y])$$(\Psi \circ \Phi)(f)([x,y]) = f_{\Phi(f)}([x,y]) = \Phi(f)(x)(y) = f([x,y])$, so $\mathrm{\Psi }\circ \mathrm{\Phi }=\mathrm{id}$$\Psi \circ \Phi =\mathrm{id}$

Naturality

One can further verify that this isomorphism is natural in $X$$X$, $Y$$Y$, and $Z$$Z$, meaning it respects morphisms in these variables.

Conclusion

We have established a natural isomorphism: ${\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X\wedge Y,Z)\cong {\mathrm{Hom}}_{{\mathrm{Set}}_{\ast }}(X,Z^Y)$$\mathrm{Hom}_{\mathrm{Set}_*}(X\wedge Y,Z) \cong \mathrm{Hom}_{\mathrm{Set}_*}(X,Z^Y)$

This proves the tensor-hom adjunction in the category of pointed sets, showing that the smash product $\wedge$$\wedge$ and the internal hom $Z^Y$$Z^Y$ form an adjoint pair of functors.

We also could define complex over ${\mathrm{Set}}_{\ast }$$\mathrm{Set}_*$.

$$⟶X_n\stackrel{d}{\to }{X}_{n-1}\stackrel{d}{\to }{X}_{n-2}⟶...$$

But what is the point?