Struggled with this homework problem for a while, would greatly appreciate some hints!

We have the following theorem (call it the universal mapping property for quotient spaces): $V,U$ are vector spaces over field $F$, $W\subset V$ a subspace. Then for every linear transformation $t:V\to U$ so that $W\subset \ker t$, there exists a unique linear transformation $T:V/W\to U$ so that $T\circ p=t$, where $p:V\to V/W$ maps elements of $V$ to their cosets, i.e. $p(v)=v+W$.

Problem setup: Let $W\subset V$ be a vector space over field $F$, $Q,U$ be vector spaces, and $\pi_Q:V\to Q$ a linear transformation with $W\subset \ker(\pi_Q)$, such that for any linear transformation $T:V\to U$ with $W\subset \ker(T)$, there exists a unique linear transformation $T_Q:Q\to U$ so that $T=T_Q\circ \pi_Q$. Prove that $Q$ is isomorphic to the quotient space $V/W$. (Hint: Use the UMP for quotient spaces 4 times!)

My attempt: I've found a linear transformation that is 1-1 from $V/W\to Q$, but am having trouble showing it's onto.

We're looking for an isomorphism mapping $V/W$ to $Q$. From applying the UMP to $\pi_Q$, we get the existence of a unique LT (linear transformation) $P_Q:V/W\to Q$ so that $P_Q\circ p=\pi_Q$. Appyling the UMP to $T$, we get a unique LT $t:V/W\to U$ so that $T=t\circ p$. (We also get some freedom to pick $T$, so it might be nice to pick $T$ surjective, since then $T=T_Q\circ \pi_Q$ is surjective, but this only shows $T_Q$ surjective, not $\pi_Q$ surjective. In any case, I'm not sure if we can do this, since $U$ may be bigger than $V$.)

We know that $\ker \pi_Q\subset W$ iff $\pi_Q$ is constant on the sets $p^{-1}(X_i)$, where $\{X_i\}_{i\in I}$ are the equivalence classes of $V/W$. So if we take $P_Q$ to be our candidate for isomorphism, we see from this fact that $P_Q$ is injective. I suspect it's onto as well, but cannot show it.

I would greatly appreciate some help either showing that my candidate will work, or another approach to finding an isomorphism. Thanks in advance!


I notice at least two things that could be causing confusion here.

The symbol $T$ is defined in one way in the first paragraph, and then in a different way in the second paragraph. Then, in the fourth paragraph, the same symbol is used again - with some latitude in its definition, but apparently referring to its use in the first paragraph.

The hint you've been given is also misleading (or so it seems to me). For, you know that the canonical surjection $p: V \to V/W$ has a universal property, and you are given that the mapping $\pi_Q: V \to Q$ has a similar universal property; but the hint apparently tells you to use the property of $p$ four times, and doesn't seem to mention the property of $\pi_Q$ at all. I would suggest using each property twice.

In deciding how to use the known and given universal properties, bear in mind that an isomorphism can be defined as a map which has an inverse. Because you are looking for an isomorphism between $V/W$ and $Q$, there could be four linear transformations in play: one from $V/W$ to $Q$ (obtained using the universal property of $p$), one from $Q$ to $V/W$ (obtained using the universal property of $\pi_Q$), and two identity transformations: one on $V/W$, and one on $Q$.

And there is nothing to stop you from applying the universal property of $p$ to the transformation $p$ itself (with $U = V/W$, in the first paragraph), and the universal property of $\pi_Q$ to $\pi_Q$ itself (with $U = Q$, in the second paragraph).

(I hope that's enough of a hint, and not too much of one.)

  • $\begingroup$ Thanks for this excellent response. I've found a 1-1 map $P_Q:V/W\to Q$, and found that its inverse $P_Q^{-1}:Q\to V/W$ is onto. When you say an isomorphism can be defined as a map which has an inverse, do we need the inverse to be 1-1? I'm having trouble showing $P_Q$ to be onto or $P_Q^{-1}$ to be 1-1, either of which would be sufficient. $\endgroup$ – manofbear Oct 3 '16 at 19:00
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    $\begingroup$ If you have found a linear transformation $Z: Q \to V/W$ such that that the composite linear transformations $Z \circ P_Q$ and $P_Q \circ Z$ are the identity linear transformations on $V/W$ and $Q$, respectively, then the fact that $P_Q$ and $Z$ are both 1-1, and both onto, follows immediately. It is a general fact, and easily proved, that if $f \circ g$ is 1-1, then $g$ is 1-1; also if $f \circ g$ is onto, then $f$ is onto. Hence, if $f \circ g$ and $g \circ f$ are both 1-1, and both onto (in particular, if they are both identity transformations), then $f$ and $g$ are both 1-1, and both onto. $\endgroup$ – Calum Gilhooley Oct 3 '16 at 19:52
  • $\begingroup$ Excellent, thanks for the quick response! I will type up what I've got as soon as I can, as you've suggested. You've been extremely helpful!! $\endgroup$ – manofbear Oct 3 '16 at 19:58

As @Calum Gilhooley as suggested, here is my solution:

  • Use the universal property of $\pi_Q$ to get the unique linear transformation $P_Q:V/W\to Q$ so that $P_Q\circ p=\pi_Q$.
  • Observe that $P_Q$ is 1-1 since $\pi_Q$ is constant on the fibers of $p$, since $W\subset \ker \pi_Q$.
  • Use the given properties of $Q$ and $\pi_Q$ with $T=p$ to get the unique LT $T_Q:Q\to U$ so that $T=T_Q\circ \pi_Q$.
  • Observe that $\pi_Q=P_Q\circ p=P_Q\circ T_Q\circ \pi_Q$, so $P_Q\circ T_Q=1_Q$. Since the identity map is bijective, $P_Q$ is onto and $T_Q$ is injective. So $P_Q$ and $T_Q$ are both isomorphisms.

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