# How to show a diagram commutes?

Let $G:c_0\rightarrow\ell_\infty,x\mapsto x$ be the inclusion map. Let $T:c_0\rightarrow {c_0}^{**},x\mapsto\hat{x}$ be the canonical embedding.

We have $\phi^*:{c_0}^{**}\rightarrow\ell_1^*,\lambda\mapsto\lambda\circ\phi$ is an isometric isomorphism and $\varphi^{-1}:{\ell_1}^*\rightarrow\ell_\infty$ is an isometric isomorphism.

I want to show the diagram below commutes. $\require{AMScd}$ $$\begin{CD} c_0 @> G > > \ell_\infty\\ @V T V V @A A \varphi^{-1}A\\ {c_0}^{**} @> >\phi^* > {\ell_1}^* \end{CD}$$

Does it suffice to show that $\varphi^{-1}\circ\phi^*\circ T$ is $G$?

• That is exactly what it means to commute. – Prince M May 12 '18 at 4:44
• It just means if you pick an element in the top left corner, it doesn’t matter which route you take to get to the top right corner. – Prince M May 12 '18 at 4:45
• Is $\phi : \ell^1 \to c_0$ the standard isometric isomorphism? Also, is $\varphi : \ell^\infty \to (\ell^1)^*$ the standard isometric isomorphism? – mechanodroid May 12 '18 at 8:55

I'm assuming that $\varphi : \ell^\infty \to (\ell^1)^*$ and $\phi : \ell^1 \to (c_0)^*$ are standard isometric isomorphisms.

Since $\varphi : \ell^\infty \to (\ell^1)^*$ is an isometric isomorphism, we can invert that arrow and show that $\varphi \circ G = \phi^* \circ T$ as functions $c_0 \to (\ell^1)^*$.

Therefore, take $x = (x_n)_n \in c_0$ and $y = (y_n)_n \in \ell^1$.

$$(\varphi \circ G)(x)(y) = \varphi(G(x))(y) = \varphi(\underbrace{x}_{\in \ell^\infty})(y) = \sum_{n=1}^\infty x_ny_n$$

because $\varphi(x)$ is a functional on $\ell^1$ acting as $y \mapsto \sum_{n=1}^\infty x_ny_n$.

On the other hand, we have

$$(\phi^* \circ T)(x)(y) = \phi^*(T(x))(y) = \phi^*(\hat{x})(y) = (\hat{x} \circ \phi)(y) = \hat{x}(\phi(y)) = \phi(y)(x) = \sum_{n=1}^\infty x_ny_n$$

because $\phi(y)$ is a functional on $c_0$ acting as $x \mapsto \sum_{n=1}^\infty x_ny_n$.

Therefore $\varphi \circ G = \phi^* \circ T$ so $G = \varphi^{-1} \circ \phi^* \circ T$.