# Quotient spaces and composition of linear maps

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.

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 $$F$$.

• 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.
– user731634
Commented Jan 16, 2020 at 12:10
• @user: Yes, since $V’\subset \mathrm{ker}\mathfrak{F}$, that means there is a unique morphism $F$ with domain $V/V’$ and defined by $F(v+V’) = \mathfrak{F}(v)$. Commented Jan 16, 2020 at 18:02
• Thanks, do you know if there is a particular theorem which tells us this?
– user731634
Commented Jan 16, 2020 at 20:55
• @user: It’s the fundamental theorem of homomorphisms. It’s valid for groups, rings, modules, and vector spaces. Commented Jan 16, 2020 at 20:56
• Thank you very much, I will look it up right away.
– user731634
Commented Jan 16, 2020 at 20:57