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

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 $$X$$?

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 $$\sigma \not\in C_1(A+B)$$.

• Thanks - so the issue is with the exactness, not with the groups itself, as I was assuming, correct?
– Anon
Commented Feb 17, 2023 at 16:37
• Yes. By definition, $C_n(A+B) \subseteq C_n(X)$, so we could compose $\psi$ with the inclusion to get a map $C_n(A)\oplus C_n(B) \to C_n(X)$, but it will not be surjective in general, so we don't get a short exact sequence. Commented Feb 17, 2023 at 16:42