Let $\mathcal{V}:=\operatorname{\textbf{FVec}}_k$ be the category of finite-dimensional vector spaces over $k$, with linear transformations as morphisms, and $\mathcal{C}:= \operatorname{\textbf{FCol}}_k$ the category of column spaces $k^n$ for all finite $n$, again with linear transformations as morphisms.

I can see why both categories are equivalent: it suffices to find an equivalence (= a fully faithful, dense (or essentially surjective)) functor $F: \mathcal{C}\to \mathcal{V}$. We will let $F$ map each object in $\mathcal{C}$ to itself, but considered as vector space, in $\mathcal{V}$. A linear map in $\mathcal{C}$ is essentially a matrix, so we can map a morphism in $\mathcal{C}$ to the corresponding linear map (i.e., itself).

  • $F$ dense? Yes: every finite-dimensional vector space $V$ over $k$ is isomorphic to a column space $k^n$, with $n=\dim V$ (note that isomorphic means that there must be an invertible morphism between $V$ and $F(k^n)=k^n$, we can take the linear map that maps vectors in $V$ to their coordinate vector in $k^n$). Question: it seems like I would have to fix a basis for $V$ in this case. Is this ok?

  • $F$ fully faithful? I worked this out myself, it essentially comes down to the fact that $F$ acts trivially on morphisms of $\mathcal{C}$.

How can we see that these categories are not isomorphic? There should be no functors $F:\mathcal{C}\to\mathcal{V}$ and $G:\mathcal{V}\to\mathcal{C}$ such that $FG=1_{\mathcal{C}}$ and $GF=1_{\mathcal{D}}$.


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    $\begingroup$ Essentially, there are "too many" vector spaces of any given dimension, while column spaces are more restricted. Although vector spaces of the same dimension are isomorphic, an isomorphism of categories requires that different-but-isomorphic-objects be mapped to different-but-isomorphic-objects, and you don't have enough of them in the category of column spaces. $\endgroup$ Apr 4, 2022 at 18:35

1 Answer 1


Zero-dimensional vector spaces are described by having a unique morphism into them from any other object, i.e they are terminal objects(and in fact zero objects as they also have unique morphism out of them).

There is exactly one zero-dimensional column space $k^0$, i.e. exactly one terminal object in the category of column spaces. Consequently, any functor that preserves terminal objects (e.g. an isomorphism, an equivalence, or a right adjoint) would have to send each zero-dimensional vector space to the zero-dimenional column space.

In particular, for such a functor to be injective on objects (as an isomorphism has to be), there should also be exaclty one terminal object in the cateogory of vector spaces, i.e. exactly one zero-dimensional space. But every singleton set $\{*\}$ has a unique structure as a zero-dimensional vector space, so the existence of multiple singletons yields multiple zero-dimensional vector spaces. Thus an injective on objects functor that preserves terminal objects (e.g. an isomorphism) does not exist.


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