# Is this isomorphism canonical?

Suppose $A\leq A',B$ and $C' \leq C$ are (finite dimensional) vector spaces. Suppose that $$0 \to A \to B \to C \to 0$$

$$0 \to A' \to B \to C' \to 0$$ are exact. Then using a dimension argument it follows $A'/A \cong C/C'$.

However, I was wondering whether there is a canonical map realising this isomorphism. By that, I mean, that after enhancing the diagram with the inclusion maps $A \to A'$, $B\to B$ and $C' \to C$ and supposing that it commutes, is there a diagram chasing argument providing the isomorphism $A'/A \cong C/C'$?

If this is the case, what is essential about vector spaces for this to work? In other words, to what categories does the argument generalize.

If this is a simple question in homological algebra/diagram chasing I would also be content with a reference for this statement.

EDIT: Following Andrews suggestion I redraw the diagram. Consider $$0 \ \ \ \to A \ \ \ \stackrel{f}{\to} \ \ \ B \stackrel{g}{\to} C \to 0$$ $$\downarrow \subseteq \ \ \ \ \ \updownarrow \cong$$ $$0 \ \ \ \to A' \stackrel{f}{\to} \ \ \ B \stackrel{g'}{\to} C' \to 0$$ So $f:A \to B$ is just the restriction of $f:A' \to B$.

As in Andrews answer there is of course a map $A' \to C \to C/C'$, and $A$ is in the kernel. From that we get a map $A'/A \to C/C'$. But why has this map to be an isomorphism? I think it does not have to be, it could just be the map sending everything to 0, right?

So is there a map which we can read off the diagram, which realizes the isomorphism $A'/A \to C/C'$?

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Wait a moment. Your two exact sequences, are they supposed to be related to each other? So is the map $A\to B$ supposed to be the restriction of the map $A^{\prime}\to B$, and the map $B\to C^{\prime}$ to be the... what? Maybe our $C^{\prime}$ should be a factor of $C$ rather than a submodule? –  darij grinberg Aug 4 '10 at 15:45
Getting off topic (but maybe this is the true topic here): What if we have, by some general constructions valid in every Abelian category, have constructed some canonical morphism well-defined in every Abelian category, and shown that it is an isomorphism whenever we are working with vector spaces (or even finite-dimensional vector spaces). Does this yield that it is an isomorphism generally, i. e. there is a canonically defined inverse morphism? –  darij grinberg Aug 4 '10 at 15:49
@darij: yes, $A \to B$ is supposed to be the restriction of the map $A' \to B$. I now see the problem, you pointed out, that the maps $B \to C'$ and $B \to C$ are actually not related, so we cannot hope for a canonical isomorphism. Thanks, I can now see my error in what I was thinking about on this problem. –  wood Aug 4 '10 at 20:36
@wood: in the edit to your question it's still not clear what the inclusion $C''to C$ is when you write $C\to C\backslash C'$. As you comment yourself, there is no natural choice for it. And if you had an "artificial" choice which made the diagram commutative then it would follow that the map $A\to C/C'$ is the zero map by commutativity. So your map sends indeed everything to zero. –  Michael Bächtold Aug 5 '10 at 16:54
I suppose that $C'$ is a fixed subspace of $C$ and that the maps $g$ and $g'$ map $B$ to $C$ and $C'$ respectively. As Andrew and others pointed out, commutativity for the right square would imply that actually $C=C'$. This is not true the setting I am looking at, so I now didn't put any arrow between $C$ and $C'$. But still $C' \leq C$. The question is: Can the isomorphism $A'/A \to C/C'$ given by a map obtained from this diagram? The proof that this is an isomorphism if not otherwise possible may use a dimension argument, but it is possible to write it down using the maps in the diagram.? –  wood Aug 5 '10 at 23:44

Okay, let's have another go. Firstly, let's observe that we may assume that $A$ is the zero space. This involves quotienting $B$ and $A'$ by $A$ but leaves $C$ and $C'$ alone. Thus our diagram is:

$$\begin{array} & && 0 & \to & B & \to & C & \to & 0 \\ & & \downarrow & & \updownarrow 1_B \\ 0 & \to & A' & \to & B & \to & C' & \to & 0 \end{array}$$

Whereupon, we get a natural isomorphism $B/A' \to C'$.

Now since every morphism in $Vect$ splits, there is a splitting map $C' \to B$ and that provides an isomorphism $B/C' \cong A'$. As $B = C$, we can rewrite that as $C/C' \cong A'$. However, as the splitting is not natural, this isomorphism is not natural and cannot be assumed to exist in an arbitrary abelian category (as far as I'm aware).

So the difficulty is not in the map $A' \to C$ - our assumption that $A = 0$ has made it clearer that that is simply the map factoring through $B$ - but the existence of the map $C' \to C$.

(As with my original answer, I'm still not sure I'm answering the question that the person is really trying to ask; I'm also aware that the abelian category stuff is not my area of expertise so I may well have overlooked something obvious. However, perhaps this will serve to focus the question yet more.)

(Original answer deleted as it was an attempt to clarify the question)

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yes, that is what I meant by "Then using a dimension argument it follows $A'/A \cong C/C'$." What I am asking is whether there is a dimension-free argument that $A'/A$ and $C/C'$ have to be isomorphic. –  wood Aug 4 '10 at 20:22
It wasn't clear that you had a map in mind. Any two vector spaces of the same dimension are isomorphic, so simply saying that the dimensions of these two is the same is enough to assert that they are isomorphic! So you need a reverse map C -> A', now I understand a little more. –  Loop Space Aug 4 '10 at 20:37
Let me rephrase the question. How do you see that $A'/A \to C/C'$ is actually an isomorphism. I mean, why could this the map you get from the diagram $A' \to C/C'$ not just map everything to 0 and not only $A$? –  wood Aug 5 '10 at 9:22
I am still somewhat unsure as to which diagrams commute and which maps go where, particularly in light of the other answers. I recommend that you carefully draw some more diagrams in your question explaining which maps you have and which diagrams are assumed to commute. –  Loop Space Aug 5 '10 at 10:17

The assumption in the question that a certain diagram commutes doesn't seem to make sense, except in trivial cases. The diagram in question consists of the two exhibited exact sequences, downward arrows $A\to A'$ and $B\to B$ and an upward arrow $C'\to C$, all representing inclusion maps (unless I'm misunderstanding something). Commutativity of the right square means that the surjection $B\to C$ factors as $B\to C'\to C$, where the $C'\to C$ part is an inclusion. That forces this inclusion to be surjective, so it's an isomorphism. Then, by dimension counting, $A\to A'$ is also an isomorphism, and the question reduces to canonicity of an isomorphism between two 0-dimensional spaces.

I conjecture that the OP didn't really intend commutativity (at least not in the sense I've understood it). But some connection between the given maps will be needed to get the desired sort of canonicity. Specifically, with no commutativity assumption about the right square (but still requiring commutativity of the left square), we get that $A'/A$ is canonically isomorphic to the kernel of a canonical projection map $C\to C'$, but to embed $C'$ into $C$ and to get an isomorphism between that kernel and $C/C'$ requires some additional structure.

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Here is an example which makes me think that the answer might be no: I work backwards from the idea that if $C=R^2$ and $C'$ is the constant vectors then $C/C'$ is isomorphic to $R$ but not in any natural way (I can think of two isomorphisms -- (x,y) goes to either x-y or to y-x -- but neither seems more natural than the other.) So I will take $A'=R$, $A=0$, $B=R^2$ and f sends x to (x,-x) while g sends (x,y) to x+y. It is true that the f map is biased, I'd like to avoid that, but haven't seen how yet.

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