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The Abelian field-strength chiral projection is linear in the vector superfield, so isolate the terms containing the gaugino $\lambda$ and the $D$ auxiliary field. Replacing $x$ by $y-i\theta \sigma \bar{\theta }$ leaves these terms unchanged: the shift of the gaugino term would contain three barred Grassmann variables, and the shift of the $D$ term would contain three of each chirality, hence both vanish. Their contribution is thereforeThe $2i({\sigma }^{\mu }\bar{\theta }{)}_{\alpha }{\partial }_{\mu }$ part of the supersymmetric covariant derivative also adds a third barred factor, so it vanishes on these two terms. The ordinary left Grassmann derivative, with ${\partial }_{\alpha }(\theta \theta )=2{\theta }_{\alpha }$, givesAt fixed $y$, $\bar{D}=-\bar{\partial }$ and ${\bar{D}}^2(\bar{\theta }\bar{\theta })=-4$. Thus the chiral field-strength superfield hasThe remaining components come from the other independent terms in the vector superfield; linearity ensures they cannot alter the two coefficients just calculated. This proves the requested components without computing the omitted terms. The gaugino phase and the sign of the vector component are those in the printed Wess-Zumino gauge expansion.

$W_{\alpha }$ is a fermionic chiral spinor superfield. Its chirality follows directly from the Abelian field-strength chiral projection:There are only two independent barred supersymmetric covariant derivatives, and their equal-chirality anticommutators vanish. Every product of three barred derivatives therefore vanishes. Equivalently, the previous calculation has no independent barred Grassmann variable at fixed $y$. The free undotted index makes this a chiral spinor superfield, rather than a scalar chiral superfield; its lowest component is the odd gaugino ${\lambda }_{\alpha }$.