From Trinity College Cambridge

@zaahir on keybase

About

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