Rotman's Algebraic Topology Lemma 6.11

I am trying to understand the proof of Lemma 6.11 of Rotman's algebraic topology book. I understood most parts but it's not obvious to me why jl = k; why does this diagram commute?

Solution 1:

Fix a non-negative integer $n$. Consider $\sum_\text{finite}c_i\cdot\sigma_i\in S_n(X_1)$, where each $\sigma_i\colon \Delta^n\to X_1$ is continuous and $c_i\in \Bbb Z$.

Now, $\ell$ sends $\sum_\text{finite}c_i\cdot\sigma_i+ S_n(X_1\cap X_2)\in \frac{S_n(X_1)}{S_n(X_1\cap X_2)}$ to $\sum_\text{finite}c_i\cdot\overline{\sigma_i}+S_n(X_2)\in \frac{S_n(X_1)+S_n(X_2)}{S_n(X_2)}$, where $\overline{\sigma_i}\colon \Delta^n\to X$ defined by $\overline{\sigma_i}(p)=\sigma_i(p)$ for all $p\in \Delta^n$, i.e., we are thinking $S_n(X_1)\subseteq S_n(X)$.

Also, we write $S_n(X_1)+S_n(X_2)$ thinking $S_n(X_1), S_n(X_2)\subseteq S_n(X)$ as defined above, i.e., $\texttt{extending co-domain}$.

Next, $j$ sends $\sum_\text{finite}c_i\cdot\overline{\sigma_i}+S_n(X_2)\in \frac{S_n(X_1)+S_n(X_2)}{S_n(X_2)}$ to $\sum_\text{finite}c_i\cdot\overline{\sigma_i}+S_n(X_2)\in \frac{S_n(X)}{S_n(X_2)}$

Finally, $k$ sends $\sum_\text{finite}c_i\cdot\sigma_i+ S_n(X_1\cap X_2)\in \frac{S_n(X_1)}{S_n(X_1\cap X_2)}$ to $\sum_\text{finite}c_i\cdot\overline{\sigma_i}+S_n(X_2)\in \frac{S_n(X)}{S_n(X_2)}$.

So, we are done.