1

Let $X$ be a topological space and $A \subset X$ a subspace. I need to prove that the inclusion $C_*(A)+C_*(X) \hookrightarrow C_*(X)$ induces an isomorphism in homology, that is, $H_q(C_*(A)+C_*(X)) \cong H_q(C_*(X))$.

($C_*$ is the singular complex).

So I was wondering if it's true that $C_*(A)+C_*(X)=C_*(X)$ since $A \subset X$. In that case, the proof would be done.

Twnk
  • 2,436
  • Something is funny about the statement of the problem. I feel like it should say $C_(A) + C_(X \setminus A)$ or something else like that. Otherwise, as you say, $C_(A) + C_(X) = C_*(X)$. – Trevor Gunn Jan 03 '20 at 03:03
  • No, the statement is correct. I know it's very simple but I'm reading a proof of a theorem and I wanna make sure I understand every datail. So if we have two R modules $Y$ and $Z$ such that $Y \subset Z$, then $Y+Z=Z$? – Twnk Jan 03 '20 at 03:07
  • 1
    Yes, $Y + Z = {y + z : y \in Y, z \in Z} \subseteq Z$ by definition, and for any $z \in Z$, $z = 0 + z \in Y + Z$ so $Z \subseteq Y + Z$. – Trevor Gunn Jan 03 '20 at 03:11
  • Do you doubt that an $n$-chain in $A$ is an $n$-chain in $X$ for any $n$? – Eric Towers Jan 03 '20 at 03:28

1 Answers1

0

So as I said in the comments: whenever you have an Abelian group (module, vector space) $A$ and subgroup $B$ then $A + B = A$ because the internal sum is just

$$ A + B = \{ a + b : a \in A, b \in B \}. $$

The internal sum of $A$ and $B$ being defined as the smallest subgroup containing both $A$ and $B$ so you can check that this meets that criterion.

So we have $A \subseteq A + B \subseteq A$ so $A + B = A$.

Trevor Gunn
  • 27,041