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Let $V,W$ be vector spaces over a field $\mathbb{F}$, $f:V\to W$ a linear map and $V'\subset V$, $W'\subset W$. Show that $f(V')\subset W'$ iff there exists a linear map $F:V/V'\to W/W'$, such that $F\circ \pi_V=\pi_w\circ f$ where $\pi_V,\pi_W$ denote the canonical linear maps to the respective quotient spaces.


I've been stuck with this for a while and would greatly appreciate any help. Thank you very much in advance.

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Suppose $F$ exists. If $v\in V’$, then $\pi_V(v)=\mathbf{0}$, so the left hand side of $F\circ \pi_V$ evaluates to $\mathbf{0}$. Hence, so does the right hand side, so $\pi_W(f(\mathbf{v}))=\mathbf{0}$. Thus, $f(\mathbf{v})\in\mathrm{ker}(\pi_W)$.

Conversely, if $f(V’)\subseteq W’$, define $\mathfrak{F}\colon V\to W/W’$ by $\pi_W\circ f$. Show that $V’\subseteq\mathfrak{F}$, so this induces the map $F$.

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  • $\begingroup$ Thank you very much for the answer, but I haven' been able to work out what you mean by that last sentence: Do you mean that $V'\subset \ker(\mathfrak{F})$? And if so, I can't really seem to understand what it means that $F$ gets induced by that fact. $\endgroup$
    – user731634
    Commented Jan 16, 2020 at 12:10
  • $\begingroup$ @user: Yes, since $V’\subset \mathrm{ker}\mathfrak{F}$, that means there is a unique morphism $F$ with domain $V/V’$ and defined by $F(v+V’) = \mathfrak{F}(v)$. $\endgroup$ Commented Jan 16, 2020 at 18:02
  • $\begingroup$ Thanks, do you know if there is a particular theorem which tells us this? $\endgroup$
    – user731634
    Commented Jan 16, 2020 at 20:55
  • $\begingroup$ @user: It’s the fundamental theorem of homomorphisms. It’s valid for groups, rings, modules, and vector spaces. $\endgroup$ Commented Jan 16, 2020 at 20:56
  • $\begingroup$ Thank you very much, I will look it up right away. $\endgroup$
    – user731634
    Commented Jan 16, 2020 at 20:57

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