External differential commutes with antipodal map Suppose we have a closed form $d\omega=0$ on $S^{n}$, and antipodal map $i: S^{n} \to S^{n}$ i.e $i : x \to -x$. How to see that the external differential commutes with antipodal map?

This is nothing special about $S^n$ or $i$.

Let $M$, $N$ be smooth manifolds. Given any smooth map $f : M \to N$, there is an induced map $f^* : \Omega^*(N) \to \Omega^*(M)$ which respects the grading, i.e. $f^* : \Omega^k(N) \to \Omega^k(M)$ for every $k$. Most books on differential geometry would contain a proof of the fact that $f^*$ commutes with $d$ (e.g. Proposition $14.26$ of Lee's _Introduction to Smooth Manifolds_ , second edition). To be specific, $d(f^*\omega) = f^*(d\omega)$ for every $\omega \in \Omega^*(N)$. This property is often called the naturality of the exterior derivative which can also be stated in terms of the commutativity of the following diagram for every $k$:

$$\require{AMScd} \begin{CD} \Omega^k(N) @>{f^*}>> \Omega^k(M)\\\ @V{d}VV @V{d}VV \\\ \Omega^{k+1}(N) @>{f^*}>> \Omega^{k+1}(M). \end{CD}$$