# Proving that the diagram commutes

Let's consider a continuous map $f: X \rightarrow Y$, so that $Y= U_{1} \cup U_{2}$ -- covering by open sets and $X = f^{-1}(U_{1}) \cup f^{-1}(U_{2}) = V_{1} \cup V_{2}$. How to prove that the following diagram commutes?


Here, $$f_{1} = f|_{V_{1}} : V_{1} \rightarrow V_{1}$$ $$f_{2} = f|_{V_{2}}: V_{2} \rightarrow V_{2}$$ $$f_{12} = f |_{V_{1} \cap V_{2}}: V_{1} \cap V_{2} \rightarrow U_{1} \cap U_{2}$$ $$\varphi: C_{n}(U_{1} \cap U_{2}) \rightarrow C_{n}(U_{1}) \oplus C_{n}(U_{2}), \varphi(x) = (x, -x)$$ $$\psi: C_{n}(U_{1}) \oplus C_{n}(U_{2}) \rightarrow C_{n}(Y), \psi(x, y) = x+ y$$

So, let's try proving that $f^{n}_{1} \oplus f^{n}_{2} \circ \varphi^{n} = \tilde{\varphi^{n}} \circ f^{n}_{12}$. So, take $x \in C_{n}(U_{1} \cap U_{2})$, then $\varphi_{n}(x) = (x, -x)$, then $f_{1}^{n} \oplus f_{2}^{n}(x, -x) = (f_{1}(x), f_{2}(-x))$, on the other hand $f_{12}(x) = f(x)$, then $\tilde{\varphi_{n}(f(x))} = (f(x), -f(x))$, so if the diagram commutes, then $f$ is odd, i.e. $f(-x)= -f(x)$. What's the problem hidden there? In general it's not required from $f$ to be an odd function.

Also, it can be shown that these long sequences are exact (it's always proven while proving that the Mayer-Vietoris sequence is exact). Does the feature of exactness will help much in this concrete case?

Any help would be much appreciated.

• In case you aren’t already on it, fixing it: You have a math processing error in your upper latex code. – k.stm Nov 22 '15 at 12:26
• @k.stm I can't see where exactly it is -- on my screen everything renders well. – hyperkahler Nov 22 '15 at 12:29