Hello I was studying the corollary to the excision property in Homotopy theory (Hatcher 4K.2) and the thing I can't understand is why the injectivity argument works when moving from an infinite covering to a finite covering.
The idea goes like this: We have a continuous function $f:X \longrightarrow Y$ and some open coverings $ \mathcal{U} = \{U_i\}_{i \in I}$ and $ \mathcal{V}=\{V\}_{i \in I}$ of $X$ and $Y$ resp. with $f(U_i) \subseteq V_i$. The objective is to prove that if the restriction to any intersection of a finite number of $U_i$'s is a weak equivalence, then $f:X \longrightarrow Y$ is a weak equivalence. You first prove the $I$ finite case which is fine and the fact that $f_*$ is an epimorphism in the infinite case which is also fine.
The issue is when you try to prove that $f_*$ is injective. Take $[g] \in \pi_n(X)$ such that $f_*([g])=0$ in $\pi_n(Y)$. By compacity $g:S^n \rightarrow X$ lies in $U_{j_1} \cup ... \cup \ U_{j_k} =: \tilde{X} $ so it factors as $g =i\circ \tilde{g}$ with $[\tilde{g}] \in \pi_n(\tilde{X})$. We have the diagram
$$ \require{AMScd} \begin{CD} \pi_n(\tilde{X}) @>\tilde{f}_*>> \pi_n(\tilde{Y})\\ @Vi_*VV @VVi_*V \\ \pi_n(X) @>f_*>> \pi_n(Y)\\ \end{CD} $$
We know, by the finite case, $\tilde{f}_*$ is injective. But the problem is that $ 0 = f_*([g]) = f_* \circ i_* ([\tilde{g}]) = i_* \circ \tilde{f} ([\tilde{g}])$ does not imply that $\tilde{f} ([\tilde{g}]) = 0$ since $i_*$ is not injective in general (think of the inclusion of the disk minus the center onto the disk) and thus we can't use the injectivity of $\tilde{f}$. Just because the curve $f(g)$ is homotopic to a point in $Y$ does not mean it is in $\tilde{Y}$.
How can we solve this problem and prove injectivity?
The idea is to actually choose the covering of the compact image of the homotopy, not the image of $g$.
$f_*([g]) =0$ which means there is $H: I \times S^n \rightarrow Y$ an homotopy between $f \circ g$ and $y_0$ relative to $y_0$. Since the image of $H$ is compact there exists $V_1, ... , V_n$ that contain it. And also since the image of $g$ is compact there exist $U_{n+1}, ... , U_R$ that contain it (in particular the corresponding $V$s contain the image of $f \circ g$). Let $\tilde{X} := U_1 \cup ... \cup U_n \cup U_{n+1} \cup ... \cup U_R$ and $\tilde{Y} := V_1 \cup ... \cup V_n \cup V_{n+1} \cup ... \cup V_R$
Then $g \subseteq \tilde{X}$, $\space$ $ f \circ g , \ H \subseteq \tilde{Y}$ and $f(g) \underset{homo.}{\cong} y_0 \ rel(y_0)$ in $\tilde{Y}$. The same homotopy whose image lies inside $\tilde{Y}$ works to prove that $\tilde{f}_*([\tilde{g}])$ must also be zero in $\pi_n (\tilde{Y})$. By the finite case $\tilde{f}_*:\pi_n(\tilde{X}) \rightarrow \pi_n(\tilde{Y})$ is an isomorphism and so $[\tilde{g}] = 0$ and finally $i_* [\tilde{g}] = [g] = 0$