# kernel, image being direct summands

• Jan 16th 2012, 07:35 PM
ymar
kernel, image being direct summands
This is embarrassing but I seem to be having trouble with the first isomorphism theorem.

I'm reading stuff about endomorphisms' kernels and images being direct summands of a module. I haven't seen any explicit statements but the authors clearly imply that whether the kernel is a direct summand is indepentent from whether the image is. I don't understand why.

If I have a direct sum of two modules $K\oplus L$ isn't it always true that $(K\oplus L)/K$ is isomorphic to $L?$ Let $\phi:[(k,l)]\longmapsto l,\,\phi: (K\oplus L)/K\longrightarrow L.$ $\phi$ is clearly an isomorphism, isn't it? I've checked twice and I can't see why it wouldn't be.

But then, for $f:M\longrightarrow M,$ if $\ker f$ is a direct summand of $M,$ say $M\cong\ker f \oplus N,$ and I get

$N\cong (\ker f \oplus N)/\ker f\cong M/\ker f \cong \mathrm{im} f$

from the first isomorphism theorem. Which would mean, in particular, that $\mathrm{im} f$ is a direct summand of $M$ too. What am I doing wrong?

(I'm sorry if it's not very readable. I hope it is but it's 4.30 am here and I barely see anything.)