Bobby Fish Is ‘All Elite’, Signs With AEW After Debut On 10/6 AEW Dynamite
Image credit: AEW
Bobby Fish is All Elite.
AEW officially signed the two-time NXT Champion following his debut on the October 6 episode of AEW Dynamite. In a tweet after the show, AEW revealed that Fish is the latest recipient of the coveted “All Elite” graphic.
Welcome to the team…@TheBobbyFish is #AllElite pic.twitter.com/7CI57sYakw
— All Elite Wrestling (@AEW) October 7, 2021
In a tweet last week, Fish challenged new AEW TNT Champion Sammy Guevara to a title match, and “The Sexy Spanish God” accepted. The bout was booked for the October 6 episode of AEW Dynamite, two months to the day of his WWE release on August 6.
On Wednesday, Guevara emerged victorious, despite Fish’s best efforts. Still, it’s clear that the unsuccessful challenger will be sticking around; he’s the newest fish in an increasingly star-studded pond.
Fish thrived throughout his WWE run; he signed with the company in 2017 and reigned supreme for the next several years as part of the Undisputed Era. After the group broke up earlier this year, Fish had a brief singles run prior to his release. Before he joined WWE, Fish also had a remarkable career in Ring of Honor, where he won the ROH World Television Championship once and the ROH World Tag Team Championship three times alongside Kyle O’Reilly.
After his WWE release, Fish was announced as a participant in MLW’s Opera Cup, which kicked off at MLW Fightland. Spoilers are available here.
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