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In Hatcher's book on P140, he said that when $n=1$, the boundary map $d_1:H_1(X^1,X^0)\rightarrow H_0(X^0)$ is the same as the simplicial boundary map $\Delta_1(X)\rightarrow\Delta_0(X)$. I know the definition of $d_n$ is $j_{n-1}\partial_n: H_n(X^n,X^{n-1})\rightarrow H_{n-1}(X^{n-1})\rightarrow H_{n-1}(X^{n-1},X^{n-2})$. So I think when $n=1$, $j$ is just identity map. But I don't understand what does author mean to identify the $d_1$ with simplicial boundary map???

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Since $H_1(X^1,X^0)=H_1^{\Delta}(X^1,X^0)$, $H_0(X^0)=H_0^{\Delta}(X^0)$ by homomorphisms induced by the canonical inclusion maps, we see that $d_1=\partial_1:H_1(X^1,X^0)\to H_0(X^0)$ is the same as the map $\partial_1^\Delta:H_1^{\Delta}(X^1,X^0)\to H_0^{\Delta}(X^0)$ in the corresponding long exact sequence of the short exact sequence of singular complexes $\Delta_n(X^0)\to\Delta_n(X^1)\to\Delta_n(X^1,X^0)$, hence we need only observe that $\partial_1^\Delta$ is the same as the simplicial boundary map.

Note that by definition $H_1^{\Delta}(X^1,X^0)$ is free abelian generated by $1$-cells of $X$ since any $1$-cell of $X$ is a relative cycle and $\Delta_2(X^1,X^0)$ is trivial. Also, $H_0^{\Delta}(X^0)$ is generated by $0$-cells of $X$, concluding that $H_1^{\Delta}(X^1,X^0)=\Delta_1(X^1)$ and $H_0^\Delta(X^0)=\Delta_0(X^0)$. It suffices to observe that $\partial_1^\Delta$ sends $1$-cells of $X$ to what the simplicial boundary map $\partial:\Delta_1(X^1)\to\Delta_0(X^1)=\Delta_0(X^0)$ sends to, which is rather straightforward by definition of $\partial_1^\Delta$: The quotient $j: \Delta_1(X^1)\to\Delta_1(X^1,X^0)$ is an isomorphism sending $1$-cells to $1$-cells and $\Delta_0(X^1)=\Delta_0(X^0)$ via the inclusion of $X^0$ into $X^1$.

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