What does it mean to say the homology of a pair is $H=\left( \mathbb{Z},1 \right)$?

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I encountered a definition which is unclear to me. When one says that the homology of $X$ is $H=\left( \mathbb{Z},1 \right)$ then I do not know what it means. I know what homology is but I do not understand what $H=\left( \mathbb{Z},1 \right)$ is.

Let $F$ be a real Banach space and $U$ a nonempty open subset. Assume $\Phi \in C^1\left(U,\mathbb{R}\right).$ Define $\Phi^d=\left\lbrace x \in E: f\left(x\right) \leq d \right\rbrace,$ then one can compute the Homology of the pair $\left( \Phi^d,\Phi^d \setminus \lbrace u_0\rbrace\right),$ $d=\Phi\left(u_0\right),$ and $u_0$ is a point of mountain pass type in the sense of Hofer. It turns out that $H=\left(\mathbb{Z},1 \right).$

These contents are contained on pages 309-310 of this paper: HOFER, H. : A note on the topological degree at a critical point of mountainpath-type, Proc . Amer. Math. Soc. 9 0 (1984), 309-315 .

Could you show me? thank you!

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This is indeed a strange notation. I have never seen anything like it before, but from context it seems that this means the homology is the graded abelian group whose only nonzero part is $\mathbb{Z}$ in degree $1$. That is, "$H=(\mathbb{Z},1)$" means that $H_i(\Phi^d,\Phi^d \setminus \lbrace u_0\rbrace)=0$ for $i\neq 1$ and $H_1(\Phi^d,\Phi^d \setminus \lbrace u_0\rbrace)=\mathbb{Z}$.