Remember that we call a -module irreducible or “simple” if it
has no nontrivial submodules. In general, an object in any
category is simple if it has no nontrivial subobjects. If a
morphism in a category has a kernel and an image — as we’ve seen
all -morphisms do — then these are subobjects of the source and
target objects. So now we have everything we need to state and
prove Schur’s lemma. Working in a category where every morphism
has both a kernel and an image, if is a morphism between two
simple objects, then either is an isomorphism or it’s the zero
morphism from to . Indeed, since is simple it has no nontrivial
subobjects. The kernel of is a subobject of , so it must either
be itself, or the zero object. Similarly, the image of must
either be itself or the zero object. If either or then is the
zero morphism. On the other hand, if and we have an isomorphism.
To see how this works in the case of -modules, every time I say
“object” in the preceding paragraph replace it by “-module”.
Morphisms are -morphism