call the maps in the first row and in the second row suppose then and thus because both are injective. so is injective.
now let then for some because is surjective. so thus because is injective. so hence
for some and therefore thus proving that is surjective.
Remark: you might have noticed that we didn't use surjectivity of the point here is that " is injective is bijective". i'll leave this, as an easy exercise, for you to prove.