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Question

The following is taken from a text which tries to proves a relation between dimensions of subspaces.

My first question comes from

the canonical injections $in_1:M\to M\oplus N$

in line 3 and 4. My understanding of direct sum $M\oplus N$ is that, when we write down this notation, $M$ and $N$ must be independent, i.e., $M\cap N=0$. But since $M$ and $N$ are two arbitrary subspaces as indicated in line 1 an 2, they are not necessarily independent. What does the author mean here? Does $\oplus$ mean something other than direct sum? Furthermore, If we assume $M\oplus N$ denotes direct sum, problems comes out immediately. In line 4, the author wrote

injections $f:M\cap N\to M\oplus N$

If $M\cap N=0$, then the function $f$ is just the trivial $0\to 0$. So is the following $g$. I believe this is not what the author intended to define.

The second question arises from the third line if Lemma 2.14. I don't really know what that line of symbols and "a short exact sequence" in the subsequent line mean. Does that line of arrows mean something arcane in the circle of math? Of course, the question may due to I don't understand what $f$ means in the first place. I wish I could understand what the author wanted to say, after I have tried hard, so could you please help me understand all these stuff? Thanks a lot.

Answer 1

There are two types of direct sums.

Given a subspaces $U_1,U_2$ of a vector space $V$ we can define their inner sum $U_1+U_2$ to be the smallest subspace of $V$ containing both $U_1$ and $U_2$. We say this inner sum is direct, if $U_1\cap U_2=0$.

Meanwhile there is the (external) direct sum of arbitrary vector spaces $U$ and $V$. It is given by their product $U\times V$ with componentwise vector space structure. The notation $U \oplus V$ is there to remind us that the direct sum of vector spaces has some very special properties, which other types of products like the product of sets don’t share.

It may be confusing to have the same symbol $\oplus$ for both the inner and external variant, but this has a reason: Up to isomorphism they are the same. Given two subspaces $U_1,U_2$ of $V$ their sum $U_1+U_2$ is internally direct if and only if it is isomorphic to the external direct sum $U_1\oplus U_2$.

Now an exact sequence is a sequence of vector spaces and homomorphisms $$...\rightarrow V_1 \overset{\phi_1}{\rightarrow} V_2 \overset{\phi_2}{\rightarrow} V_3 \rightarrow...$$ with the property $\operatorname{Im} \phi_n = \operatorname{ker} \phi_{n+1}$ for all $n$. I cannot go into details here but they are very useful. Anyway it is a good exercise in the definitions to check that for subspaces $U_1, U_2$ of $V$ the sequence $$ 0 \rightarrow U_1 \cap U_2 \rightarrow U_1 \oplus U_2 \rightarrow U_1+U_2 \rightarrow 0$$ proposed by your author is exact.

Answer 2

Leaving this as an answer because it might be too long for a comment.

You might be confused by what it means to be a direct sum. A direct sum $M \oplus N$ is as in Wikipedia. What you might be referring to is that $M + N = M \oplus N = V$ if and only if $M \cap N = \mathbf{0}$ the zero subspace. See for example this older MSE question. There is often a bit of trouble in the terminology if you're unfamiliar with it, since the direct sum is 'essentially' the direct product (for a finite collection of vector spaces - for an infinite collection, a direct sum is usually imposed by a 'local finiteness' condition, that in any sum there are only finitely many nonzero coefficients). This definition should answer your first two questions.

As for the second question, a short exact sequence is defined here in Wikipedia. It is a sequence of five terms, two of which are zero, and it is exact at each component (it's probably helpful to understand what the definition of 'exactness' means - again, look at Wikipedia).

In the setting of your given lemma, you have injections $f, g: M \cap N \to M \oplus N$ coming from the canonical inclusions $M, N \hookrightarrow M + N$ (can you see why this is the case? This is very straightforward, spelled out in the text, and also follows from our definition above). Once you've understood this, hopefully the statement should make sense.