Introduction E9F5FC

Questions FFFFC0

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Category theory

Read p.43-60. Exercises 6, 7, 23.

A monic morphism {$f:X\rightarrow Y$} is a left-cancellative morphism, which is a generalization of an injective homomorphism. For all objects {$Z$} and morphisms {$g_1,g_2:Z\rightarrow X$} we have {$f\circ g_1 = f\circ g_2 \Rightarrow g_1=g_2$}. Equivalently, {$g_1 \neq g_2 \Rightarrow f\circ g_1 \neq f\circ g_2 $}.

Functor {$\textrm{hom}_C(A,\cdot)$} maps {$\alpha:C\rightarrow D$} to the set function {$\alpha^\leftarrow :\textrm{hom}_\mathbf{C}(A,C)\rightarrow \textrm{hom}_\mathbf{C}(A,D)$} which sends {$h:A\rightarrow C$} to {$\alpha \circ h:A\rightarrow D$}, which is to say, {$\alpha^\leftarrow(h)=\alpha \circ h$}.

Suppose {$\alpha:C\rightarrow D$} is monic.

Suppose we are given set functions {$g_1, g_2:Z\rightarrow \textrm{hom}_\mathbf{C}(A,C)$} such that {$g_1\neq g_2$}.

Then {$Z$} must be nonempty. There exists {$z\in Z$} such that {$g_1(z)\neq g_2(z)$}. Note that {$g_1(z),g_2(z)\in \textrm{hom}_\mathbf{C}(A,C)$}.

{$\alpha$} is monic and therefore {$\alpha \circ g_1(z) \neq \alpha \circ g_2(z)$}.

Note that for {$i=1,2$} we have {$(\alpha^\leftarrow \circ g_i)(z)=\alpha^\leftarrow(g_i(z))=\alpha \circ g_i(z)$}.

Consequently, there exists {$z\in Z$} such that {$(\alpha^\leftarrow \circ g_1)(z)=\alpha^\leftarrow(g_1(z))=\alpha \circ g_1(z) \neq \alpha \circ g_2(z) = \alpha^\leftarrow(g_2(z))=(\alpha^\leftarrow \circ g_2)(z)$}.

Thus {$(\alpha^\leftarrow \circ g_1)(z) \neq (\alpha^\leftarrow \circ g_2)(z)$}.

Therefore {$\alpha^\leftarrow$} is monic.

Suppose we have an arrow {$f:X\rightarrow Y$} in {$\mathbf{C}$}.

The hom functor takes {$f$} to the set function {$\theta : \textrm{hom}_\mathbf{C}(B,X)\rightarrow \textrm{hom}_\mathbf{C}(B,Y)$} defined as follows.

Given {$g:B\rightarrow X$}, {$\theta (g)=f\circ g = B\overset{g}{\rightarrow} X \overset{f}{\rightarrow} Y$}. Which is to say, {$f\circ g$} extends {$g$} by {$f$}.

Note that {$g\in \textrm{hom}_\mathbf{C}(B,X)$} and {$f\circ g\in \textrm{hom}_\mathbf{C}(B,Y)$}.

Naturalness

Suppose we have two such hom functors {$\textrm{hom}_\mathbf{C}(A,\cdot)$} and {$\textrm{hom}_\mathbf{C}(B,\cdot)$}.

Any arrow {$g:B\rightarrow A$} defines a natural transformation {$\eta(g):\textrm{hom}_\mathbf{C}(A,\cdot)\rightarrow \textrm{hom}_\mathbf{C}(B,\cdot)$}.

Given any arrow {$f:X\rightarrow Y$} in {$\mathbf{C}$}, the naturalness condition is that the following diagram is commutative:

{$$ \matrix{ \textrm{hom}_\mathbf{C}(A,X) & \overset{\eta(g)_X \textrm{ prepends } g:B\rightarrow A}{\longrightarrow} & \textrm{hom}_\mathbf{C}(B,X) \cr \scriptsize\textrm{hom}_\mathbf{C}(A,\cdot)(f) \textrm{ postpends } f:X\rightarrow Y \bigg\downarrow & & \bigg\downarrow \scriptsize\textrm{hom}_\mathbf{C}(B,\cdot)(f) \textrm{ postpends } f:X\rightarrow Y \cr \textrm{hom}_\mathbf{C}(A,Y) & \underset{\eta(g)_Y \textrm{ prepends } g:B\rightarrow A}{\longrightarrow} & \textrm{hom}_\mathbf{C}(B,Y) \cr } $$}

This simply means that given {$h:A\rightarrow X$}, we can extend it by prepending {$g:B\rightarrow A$} and postpending {$f:X\rightarrow Y$}, and it doesn't matter which we do first. We will get {$f\circ h\circ g = B \overset{g}{\rightarrow} A \overset{h}{\rightarrow} X \overset{f}{\rightarrow} Y$}.

The functor {$H$} maps the arrow {$F(C)\overset{\lambda_C}{\rightarrow}G(C)$} in {$\mathbf{D}$} to the arrow {$HF(C)\overset{H\lambda_C}{\rightarrow}G(C)$} in {$\mathbf{E}$}.

Given an arrow {$f:A\rightarrow B$} in {$\mathbf{C}$}, then we have the following commutative diagram in {$\mathbf{D}$} because {$\lambda:F\rightarrow G$} is a natural transformation:

{$$ \matrix{ F(A) & \overset{\lambda_A}{\longrightarrow} & G(A) \cr \scriptsize F(f) \bigg\downarrow & & \bigg\downarrow \scriptsize F(f) \cr F(B) & \underset{\lambda_B}{\longrightarrow} & G(B) \cr } $$}

Applying the functor {$H$} to this entire commutative diagram, we get the following commutative diagram in {$\mathbf{E}$}:

{$$ \matrix{ HF(A) & \overset{H\lambda_A}{\longrightarrow} & HG(A) \cr \scriptsize HF(f) \bigg\downarrow & & \bigg\downarrow \scriptsize HF(f) \cr HF(B) & \underset{H\lambda_B}{\longrightarrow} & HG(B) \cr } $$}

Thus the family {$H\lambda$} is a natural transformation from the functor {$FH$} to the functor {$FG$}.

Given an arrow {$g:X\rightarrow Y$} in {$\mathbf{B}$}, consider the diagram below. Is it commutative?

{$$ \matrix{ FK(X) & \overset{\lambda_{KX}}{\longrightarrow} & GK(X) \cr \scriptsize FK(g) \bigg\downarrow & & \bigg\downarrow \scriptsize GK(g) \cr FK(Y) & \underset{\lambda_{KY}}{\longrightarrow} & GK(Y) \cr } $$}

Simply introduce parentheses in this diagram to emphasize that the objects and arrows {$K(g):K(X)\rightarrow K(Y)$} are in {$C$}.

{$$ \matrix{ F(K(X)) & \overset{\lambda_{K(X)}}{\longrightarrow} & G(K(X)) \cr \scriptsize F(K(g)) \bigg\downarrow & & \bigg\downarrow \scriptsize G(K(g)) \cr F(K(Y)) & \underset{\lambda_{K(Y)}}{\longrightarrow} & G(K(Y)) \cr } $$}

{$\lambda:F\rightarrow G$} is a natural transformation, and so this is a commutative diagram in {$\mathbf{D}$}. By the same diagram, {$\lambda K:FK\rightarrow GK$} is a natural transformation.

The natural transformation {$\alpha$} has components {$\alpha_A:F(A)\rightarrow G(A)$} indexed by objects {$A$} in {$\mathbf{C}$}. Such a component is a morphism in {$\mathbf{B}$}. Apply to it the natural transformation {$\beta :H\rightarrow K$}. This yields the commutative diagram:

{$$ \matrix{ H(F(A)) & \overset{\beta_{F(A)}}{\longrightarrow} & K(F(A)) \cr \scriptsize H\alpha_A \bigg\downarrow & & \bigg\downarrow \scriptsize K\alpha_A \cr H(G(A)) & \underset{\beta_{G(A)}}{\longrightarrow} & K(G(A)) \cr } $$}

This yields {$K\alpha_A\circ\beta_{F(A)}=\beta_{G(A)}\circ H\alpha_A$} and simply note that {$F(A)=FA$} and {$G(A)=GA$}.

I will write the product like this: {$(\beta * \alpha)_A:=K\alpha_A\circ\beta_{F(A)}=\beta_{G(A)}\circ H\alpha_A$} to emphasize that {$A$} is an object and {$F$} is a functor.

Given arrow {$f:X\rightarrow Y$} in {$\mathbf{A}$}, we wish to show that the following is a commutative diagram:

{$$ \matrix{ HF(X) & \overset{(\beta * \alpha)_X}{\longrightarrow} & KG(X) \cr \scriptsize HF(f) \bigg\downarrow & & \bigg\downarrow \scriptsize KG(f) \cr HF(Y) & \underset{(\beta * \alpha)_Y}{\longrightarrow} & KG(Y) \cr } $$}

Let {$\psi_f:F(X)\rightarrow G(Y)$} be the morphism in {$\mathbf{B}$} which demonstrates that {$\alpha$} is a natural transformation, namely {$\psi_f = \alpha_Y\circ F(f) = G(f)\circ\alpha_X$}.

Then {$(\beta * \alpha)_Y\circ HF(f) = \beta_{G(Y)}\circ H\alpha_Y\circ HF(f) = \beta_{G(Y)}\circ H(\alpha_Y\circ F(f)) = \beta_{G(Y)}\circ H(\psi_f)$}.

And {$KG(f) \circ (\beta * \alpha)_X = KG(f) \circ K\alpha_X\circ\beta_{F(X)} = K(G(f) \circ \alpha_X)\circ\beta_{F(X)} = K(\psi_f)\circ \beta_{F(X)}$}.

Since {$\beta:H\rightarrow K$} is a natural transformation, this is a commutative diagram:

{$$ \matrix{ H(F(X)) & \overset{\beta_{F(X)}}{\longrightarrow} & K(G(X)) \cr \scriptsize H(\psi(f)) \bigg\downarrow & & \bigg\downarrow \scriptsize K(\psi(f)) \cr H(F(Y)) & \underset{\beta_{G(Y)}}{\longrightarrow} & K(G(Y)) \cr } $$}

Which proves the desired equality.

If {$H=K$}, then we have {${id}_H * \lambda = H\lambda$}.

If {$F=G$}, then we have {$\lambda * {id}_F = \lambda F$}.