Why $A+B$ instead of $X$ required for Mayer Vietoris sequence construction?

Question

I'm wondering why, in the construction of the Mayer-Vietoris sequence in Hatcher's Algebraic Topology (p. 149), we must go through $C_n(A+B)$ instead of $C_n(X)$:

Specfically, why does $\psi(x,y)=x+y$ not provide a singular $n$-simplex in $C_n(X)$? I've been trying to think of a counter example where $\sigma_1 : \Delta^n \to A$ and $\sigma_2 : \Delta^n \to B$ do not yield an $n$-chain $\sigma_1 +\sigma_2$ of the form $\sum_i n_i \sigma_i$ where $\sigma_i : \Delta^n \to X$ are continuous. I tried using $X=S^0$, but it seems that this space is too simple to yield a counter example (it seems to work out).

Any ideas?

Also, what if we allow the choice of a different map to $\psi$? Could we then replace $A+B$ with $X$?

Answer 1

The group $C_n(A+B)$ is defined to be the subgroup of $C_n(X)$ consisting of chains that are sums of chains in $A$ and chains in $B$. So the map $\psi : C_n(A)\oplus C_n(B) \to C_n(A+B)$ given by $\psi(x, y) = x + y$ is clearly surjective by the definition of $C_n(A+B)$. This would no longer be the case if we replaced the codomain by $C_n(X)$.

For example, if $X = [0, 2]$ with $A = [0, 1.1)$ and $B = (0.9, 2]$, then the chain $\sigma : \Delta^1 \to X$ given by $\sigma(t) = \frac{1}{2}(t+1)$ has $\sigma(\Delta^1) = [\frac{1}{2}, \frac{3}{2}]$ which is not contained in $A$ or $B$, so $\sigma \not\in C_1(A+B)$.