To show that a natural transformation is an isomorphism

Question

Given a natural transformation $\tau$ from a functor $D$ to a functor $E$ where each functor is from a category $G$ to a category $C$, in order to show that if each $\tau_x$ is an isomorphism, then $\tau$ is an isomorphism, one needs to show that $\tau_y{D(u)} = E(u)\tau_x$ implies $\tau_y^{-1}E(u) = D(u)\tau_x^{-1}$ for all $x,y$ in $G$. So I managed to show that $\tau_y{D(u)} = E(u)\tau_x$ implies $\tau_y^{-1}E(u) = D(u)\tau_x^{-1}$ for all $x,y$ in $G$. But how does this prove that $\tau$ is an isomorphism?

A natural transformation $\tau:D \to E$ is defined as the collection of all morphisms $\tau_x:D(x) \to E(x)$ where $x$ is an object in $G$.

Any help will be greatly appreciated.

Answer 1

I think your problem is about what is an invertible natural transformation. By definition of an isomorphism in the category $[G,C]$ (the category of functors from $G$ to $C$), a natural transformation $\tau : D \to E$ is invertible if and only if it exists $\sigma: E \to D$ such that $\tau\sigma = \mathrm{id}_E$ and $\sigma\tau = \mathrm{id}_D$. The claim is that if each of the $\tau_x$ are isomorphisms in $E$, then the collection $$ \left({\tau_x}^{-1}:E(x)\to D(x) \right)_{x\in G} $$
is a good candidate for the wanted $\sigma$.

What you already proved is that indeed this collection is a natural transformation. Now we need to prove that $\tau\sigma = \mathrm{id}_E$ and $\sigma\tau = \mathrm{id}_D$. What is the composite of two natural transformations? By definition it is the collection of the component-wise composite. So in our case we end up with: $$ \tau\sigma = \left(E(x) \overset {{\tau_x}^{-1}} \to D(x) \overset{{\tau_x}} \to E(x)\right)_{x\in G} \quad \text{and} \quad \sigma\tau = \left(D(x) \overset {{\tau_x}} \to E(x) \overset{{\tau_x}^{-1}} \to D(x)\right)_{x\in G} $$

I let you take it from here!

Answer 2

You want to prove that the following square commutes:

\begin{array}{c} Ex & \overset{\tau_x^{-1}}\rightarrow & Dx \\ Ff\downarrow && \downarrow Gf \\ Ey & \underset{\tau_y^{-1}}\rightarrow & Dy \end{array}

Now, $Gf\circ \tau_x^{-1}=\tau_y^{-1}\circ Ff\Leftrightarrow Gf=\tau_y^{-1}\circ Ff\circ \tau_x\Leftrightarrow \tau_y\circ Gf=Ff\circ \tau_x,\ $ which is a naturality square for $\tau,\ $ which is known to commute:

\begin{array}{c} Dx & \overset{\tau_x}\rightarrow & Ex \\ Gf\downarrow && \downarrow Ff \\ Dy & \underset{\tau_y}\rightarrow & Ey \end{array}