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Let $V,W$ and $Y$ vector spaces over $F$ and let $f\in$ Hom($V$,$W$). Prove that the set of every $g \in$ Hom($W$,$Y$) such that $g\circ f=0$ is a subspace of Hom($W$,$Y$), calculate its dimension.

So I've got tackled down the subspace part, but for the dimension part I'm a bit stuck. I already got a formula for the dimension, if $A$ is the set of all $g$ such that $g\circ f=0$, then:

$$\dim(A)=\dim(\text{Hom}(W,Y)) - \dim(\text{Hom}(\text{Im}f,Y))$$

As far as I got it this is because the image of $f$ has a set $B$ such that $f \oplus B=W$ so $\dim(\text{Hom}(W,Y)) = \dim(\text{Hom}(\text{Im}f,Y))+\dim(\text{Hom}(B,Y))$. So you have to prove that $\dim(\text{Hom}(B,Y))= \dim(A)$. But how can you do that? I know you can establish a bijection between those two sets, but I'm not sure of how.

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  • $\begingroup$ Hint: consider the projection $\pi:W\to B$. $\endgroup$
    – Eero Hakavuori
    Mar 30, 2017 at 19:14
  • $\begingroup$ Well this would be $\pi(w)=b+0=b+g(f(v))$, right? So the range of this function would be the dimension of $B$. The kernel of $\pi$ is $\text{Im}(f)$, isn't it, so should we apply $g$ to $\pi$ or am I too lost in here? @EeroHakavuori $\endgroup$
    – user371816
    Mar 31, 2017 at 5:51

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