From Trinity College Cambridge
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$$\displaystyle {\displaystyle \qquad \dots \xrightarrow {\partial } {\tilde {h}}_{n}(A)\xrightarrow {i_{*}} {\tilde {h}}_{n}(X)\xrightarrow {q_{*}} {\tilde {h}}_{n}(X/A)\xrightarrow {\partial } {\tilde {h}}_{n-1}(A)\xrightarrow {i_{*}} \dots }$$