# Is there a dumbed down version of the open mapping theorem for finite dimensional real vector spaces?

I would like to understand why any surjective linear transformation between finite dimensional topological real vector spaces, each with the natural topology, is in fact an open map.

Reading around, I think this follows as a special case of the open mapping theorem from functional analysis. However, the theorems I've found, namely Theorem 2.11 in Rudin's Functional Analysis and Theorem 11.4 in Kelley & Namioka, are a little over my head at this point.

Is there a less general version of the proof specifically for finite dimensional $\mathbb{R}$-vector spaces? I have a fair understanding of general topology, real-analysis, and linear algebra, but not so much functional analysis. Thanks.

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I would post this as a comment, since I need to check it a little more before I am totally ok with it--alas, it is too long. So, an answer it shall stay.

EDIT: I should emphasize that where I am using finite dimesionality is that every linear map in sight is continuous. In particular, a linear isomorphism is a homeomorphism.

I think this should suffice. Let $V$ and $W$ be finite dimensional real vector spaces and $T:V\to W$ a surjective linear map.

EDIT EDIT: As Matt E points out below, it is prudent to mention the following. Below I prove that the map $\pi$ is open when we give $V/Y$ the quotient topology, but not that it is open when we give it the standard metric topology which makes $\widetilde{T}$ a homeomorphism. But, the fact that the quotient topology on $V/Y$ coincides with the linear topology is deducible because of the following: a) we're in finite dimensions so $Y$ is closed, and b) give $V$ you're favorite norm, since $Y$ is closed it's ommonly known then that the quotient topology on $V/Y$ is the topology induced by the quotient norm and so the quotient topology is the linear topology (all normed topologies are the same in finite dimensions). There are other ways to see that the two topologies coincide if you're interested. :)

We first claim that the projection map $V\to V/Y$ is open for any $Y$ a subspace of $V$. To see this, we note that it suffices to prove that $\pi^{-1}(\pi(U))$ (where $\pi$ is the projection map) is open for each open set $U$ in $V$. But, note that $$\pi^{-1}(\pi(U))=\bigcup_{v\in Y}(v+U)$$

which, being the union of open sets, is open.

Now that we have the projection map is open this should be easy. Namely, let $Y=\ker T$. We know then that $T$ induces a linear isomorphism $\widetilde{T}:V/Y\to W$. Since this is a linear isomorphism, it is, in particular, a homeomorphism and so an open map. So, if we take $U\subseteq V$ open we have that $T(U)$ is just $\widetilde{T}(\pi(U))$ where $\pi$ is the projection $V\to V/Y$. So, we see that $\pi(U)$ is open in $V/Y$ since $\pi$ is an open map, and $\widetilde{T}(\pi(U))$ is open because $\widetilde{T}$ is a homeomorphism. So, $T(U)=\widetilde{T}(\pi(U))$ is open as desired.

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Thanks Alex. I think I understand everything in your last paragraph. However, I don't get how you proved the projection $V\to V/Y$ is open. Why does it suffice to show $T^{-1}(T(U))$ is open, and why does the equality $$T^{-1}(T(U))=\bigcup_{v\in\ker Y}(v+U)$$ hold? Is $T^{-1}(T(U))\subset V$, but $\bigcup (v+U)\subset V/U$? – Noomi Holloway Dec 16 '12 at 6:31
@NoomiHolloway I made a typo--see if it makes more sense now that I've edited. – Alex Youcis Dec 16 '12 at 6:33
Thanks! Should the union be over $v\in\ker\pi$? If so, I understand the equality. Do you mind explaining why it suffices to show $\pi^{-1}(\pi(U))$ is open for each open $U$ in $V$ to prove $\pi\colon V\to V/Y$ is open? – Noomi Holloway Dec 16 '12 at 6:37
@Noomi: Dear Alex and Noomi, The definition of $\pi$ being open is that $\pi(U)$ is open. So the claim being made is (equivalent to the claim) that a subset of $V/Y$ is open iff its preimage in $V$ is open, or equivalently, that $V/Y$ is equipped with quotient topology induced by regarding it as a quotient of $V$. This is true; to see it, it might be easiest to choose a basis for $V$ extending a basis for $Y$, so that $V$ can be regarded as the product of $Y$ and $V/Y$. Regards, – Matt E Dec 16 '12 at 6:41
@MattE Thanks for that input! While I see how the approach you mention works, I am confused whether or not you are saying that I made a mistake. – Alex Youcis Dec 16 '12 at 6:48

Any finite-dimensional (edit: Hausdorff) topological real vector space has the usual Euclidean topology (exercise), and any surjective linear transformation between finite-dimensional topological real vector spaces is isomorphic to projection to some subset of coordinates on $\mathbb{R}^n$ (exercise), so this follows from the fact that any projection of an open Euclidean ball is a (possibly lower-dimensional) open Euclidean ball.

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