Obtaining mapping cone complex from $f:X\to Y$

This is related to Hatcher, Algebraic Topology Cor 3A.7(b).

Cor 3A.7(b) $$f:X\to Y$$ induces integral homology isomoprhism iff it induces isomorphism with coefficients over $$Q$$ and for all $$Z_p$$. "Statement (b) follows by mapping cone."

My consideration is the following. $$f:X\to Y$$. Denote cone $$X$$ as $$CX$$. Set the tip of $$CX=(X\times I)/(X\times 1)$$ as $$z=X\times 1\in CX$$. Then mapping cone $$C(f)=CX\cup Y$$ where $$\cup$$ is identified under $$f$$ mapping at the bottom $$X\times 0$$.

Consider $$A=X\times [1/2,1]\subset CX$$ and $$B=X\times [0,1/2]\cup Y$$. Clearly $$A\cap B=X\times\{1/2\}$$. Now $$B$$ deformation retracts to $$Y$$ and $$A$$ deformation retracts to $$z$$. Consider the following chain complex to induce mayer-vietoris sequence. Note that $$A$$ deformation retracts to a point implies that reduced homology is trivial.

$$0\to C_\star(A\cap B)\to C_\star(A)\oplus C_\star(B)\to C_\star(A\cup B)\to 0$$

For all reduced homology, this is equivalent to $$\dots H_\star(X)\to H_\star(Y)\to H_\star(C(f))\to H_{\star-1}(X)\dots$$ sequence. Then it follows $$H_\star(C(f),Z)=0$$ iff $$H_\star(C(f),Q)=0$$ and $$H_\star(C(f),Z_p)=0$$ for all $$p$$.

$$\textbf{Q:}$$ Normally, one would construct $$f:C(X)\to C(Y)$$ and define $$C(f)=C(Y)\oplus C(X)[-1]$$. Then this gives short exact sequence $$0\to C(Y)\to C(f)\to C(X)[-1]\to 0$$. How would I construct such a thing without knowing existence of simplicial theory say starting point is singular complex only?

For a pair of subspaces $$A, B \subset X$$ such that $$X$$ is the union of the interiors of $$A$$ and $$B$$, this exact sequence has the form $$\dots \to H_n(A \cap B) \stackrel{\Phi}{\rightarrow} H_n(A) \oplus H_n(B) \stackrel{\Psi}{\rightarrow} H_n(X) \stackrel{\partial}{\rightarrow} H_{n-1}(A \cap B)\to \dots \to H_0(X) \to 0$$
Hatcher's proof also shows how $$\Phi, \Psi$$ look like. In fact, if $$i_A : A \cap B \to A, i_A : A \cap B \to B, j_A : A \to X, j_B : B \to X$$ are the inclusion maps, then $$\Phi(\xi) = ((i_A)_*(\xi),-((i_B)_*(\xi))$$ and $$\Psi(\eta_A,\eta_B) = (j_A)_*(\eta_A) + (j_B)_*(\eta_B)$$.
Let $$p : X \times I \sqcup Y \to C(f)$$ denote the quotient map. Let $$A = p(X \times [0,1) \sqcup Y)$$ and $$B = p(X \times (0,1])$$. These are open subspaces of $$X$$ whose union is $$C(f)$$. $$B$$ is contractible and $$A$$ strongly deformation retracts to $$Y$$ (via $$r : A \to Y$$). We have $$A \cap B = p(X \times (0,1))$$ and $$\tilde{H}_n(B) = 0$$. The "projection" $$\pi : p(X \times (0,1)) \to X$$ is a homotopy equivalence and you see that $$f \pi \sim r i_A$$. This gives you an exact sequence $$\dots \to \tilde{H}_n(X) \stackrel{f_*}{\rightarrow} \tilde{H}_n(Y) \to \tilde{H}_n(C(f)) \to \tilde{H}_{n-1}(X) \to \dots \to \tilde{H}_0(C(f)) \to 0$$ Now you can apply part (a) of Corollary 3A.7.