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Let $ f: U \to V $, $g: V \to W $. Proof that $ Ker(f) \le Ker(gf)$ and

$Im(gf)\ge Im(g)$.

Intuitively it's obvious, I tried to draw a schema but I don't know how to prove it properly.

Maybe I could somehow use that: $ dimIm(gf) + dimKer(gf) = V $

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    As subsets? V has finite dim? – 2017-01-28
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    I think all of those vector spaces have finite dimensions. – 2017-01-30

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Well, I think I might came up with the solution for the image.

For the image inequality we have the equality if the $f$ is at least epimorphism (better isomorphism) because then the image of $f$ has the dimension of whole $V$ and the mapping from $V$ to $W$ (I mean for g(f)) would be the same as just that linear transformation given by $g$.

But if $f$ wasn't epimorphism, then the linear map from $V$ to $W$ would have smaller "pattern" so the image of $g(f)$ would be also potentially smaller. So the linear map $g$ could have the bigger "pattern" because we can use whole V (if this mapping is at least monomorphism) as the "pattern". In this case we could have sharp inequality.

Sorry for my english, it's hard for me.