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I have a very easy question but I can't find the solution.

Let $V,U,W$ be three $\mathbb{R}$-vector spaces and let $f: V \rightarrow U$ be a surjective linear map and $g: V \rightarrow W$ a linear map. Now, define $h : U \rightarrow W$ such that $h \circ f =g$.

I want to know if $h$ is linear or not.

Let $u_1,u_2 \in U$ such that $u_1= f(v_1)$ and $u_2= f(v_2)$ and let $c \in \mathbb{R}$. Then,

\begin{equation} \begin{aligned} h(cu_1+u_2) &= h(cf(v_1)+f(v_2)) \\ &= h(f(cv_1+v_2)) \\ &=g(cv_1+v_2) \\ &=cg(v_1)+g(v_2) \\ &= c h(f(v_1))+h(f(v_2)) \\ &= c h(u_1) +h(u_2) \end{aligned} \end{equation}

Then, I want to conclude than $h : U \rightarrow W$ is linear.

But now, take $f$ and $g$ such that $\text{ker} (f) \neq \text{ker} (g)$ (and suppose the kernel are not reduce to $0$). Then, there exists $v \in V$ such that $v \in \text{ker} (f)$ and $v \not\in \text{ker} (g)$ and so

$$h(0) = h(f(v))=g(v) \neq 0$$ and so $h$ is not linear ...

Can you explain to me what the problem is with my reasoning? My first calculation tells me that in any case $h$ is linear but the second says that when $\text{ker} (f) \neq \text{ker} (g)$ then $h$ is not linear ...

Anne Bauval
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eomp
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  • Yes, indeed, thank you for this remark (even if in my problem $f$ and $g$ are interchangeable). – eomp Jan 09 '24 at 13:19
  • I meant interchangeable in the sense that if you take your case in which $\exists v \in V$ such $v \in \text{ker }g$ and $v \not\in \text{ker }f$ then the same problem happens with the function $h' : W \rightarrow V$ defined by $h' \circ g =f$ (suppose $g$ is surjective if you want) – eomp Jan 09 '24 at 13:26
  • As Anne Bauval said (implicitly), in your case $\operatorname{ker}f\subseteq\operatorname{ker}g$. – Antonio Jan 09 '24 at 13:28

1 Answers1

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Your proof is correct: $h$ is necessarily linear.

In your (attempt of a) counterexample, you assumed $\ker f\ne\ker g$ and "derived" the existence of some $v \in V$ such that $v \in\ker f$ and $v \notin\ker g$. It ain't necessarily so . There could as well exist some $v \in V$ such that $v \in\ker g$ and $v \notin\ker f$.

Actually, this is the only possible case, because your hypothesis $h\circ f=g$ implies that $$\forall x,y\in V\quad\left(f(x)=f(y)\implies g(x)=g(y)\right),$$ in particular (special case $y=0$) $$\ker f\subseteq\ker g.$$ So, $\ker f\ne\ker g$ iff $\ker f\subsetneqq\ker g$, and the existence of a vector $v$ in $\ker g\setminus\ker f$, i.e. such that $f(v)\ne0$ and $h(f(v))=0$, does not prevent $h$ from being linear (it will only force its kernel not to be reduced to $\{0\}$).

Anne Bauval
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