[proofplan] We show that the cone $K$ is chain homotopy equivalent to the one-point complex $\{v_0\}$, from which the claimed homology groups follow immediately. The retraction $r: K \to \{v_0\}$ and the inclusion $i: \{v_0\} \hookrightarrow K$ induce chain maps, and we have $r_\bullet \circ i_\bullet = \operatorname{id}$ on the nose. The substance of the proof is the construction of an explicit chain homotopy $h: C_\bullet(K) \to C_{\bullet+1}(K)$, defined by coning oriented simplices onto $v_0$, and the verification of the chain homotopy identity $d \circ h + h \circ d = \operatorname{id} - i_\bullet \circ r_\bullet$. Once the chain homotopy equivalence is established, invariance of homology under chain homotopy equivalence transfers the homology of the point to the homology of $K$. [/proofplan]

[step:Set up the cone data and the candidate chain homotopy equivalence] By hypothesis $K$ is a cone with cone point $v_0$: every simplex of $K$ either contains $v_0$, or, if $\sigma = [w_0, \dots, w_p]$ is a simplex of $K$ not containing $v_0$, then $[v_0, w_0, \dots, w_p]$ is also a simplex of $K$. We write $(v_0, \sigma) := [v_0, w_0, \dots, w_p]$ for this coning construction on oriented simplices. Let $\{v_0\}$ denote the simplicial complex with a single $0$-simplex, whose chain complex has $C_0 = \mathbb{Z}\langle v_0 \rangle$ and $C_n = 0$ for $n > 0$. Consider the simplicial maps \begin{align*} i: \{v_0\} &\hookrightarrow K, & r: K &\to \{v_0\} \\ v_0 &\mapsto v_0, & w &\mapsto v_0 \text{ for every vertex } w \text{ of } K. \end{align*} These are simplicial maps ($r$ collapses every simplex of $K$ to the vertex $v_0$, which is a simplex of the target), so by [Simplicial Maps Induce Chain Maps](/theorems/1926) they induce chain maps $i_\bullet: C_\bullet(\{v_0\}) \to C_\bullet(K)$ and $r_\bullet: C_\bullet(K) \to C_\bullet(\{v_0\})$. Observe that $r \circ i = \operatorname{id}_{\{v_0\}}$ as simplicial maps, and hence $r_\bullet \circ i_\bullet = \operatorname{id}_{C_\bullet(\{v_0\})}$ as chain maps. [/step]

[step:Define the coning chain homotopy $h: C_\bullet(K) \to C_{\bullet+1}(K)$]For each $n \geq 0$, define $\mathbb{Z}$-linear maps \begin{align*} h_n: C_n(K) &\to C_{n+1}(K) \\ [w_0, \dots, w_n] &\mapsto \begin{cases} [v_0, w_0, \dots, w_n] & \text{if } v_0 \notin \{w_0, \dots, w_n\}, \\ 0 & \text{if } v_0 \in \{w_0, \dots, w_n\}. \end{cases} \end{align*} Extend $\mathbb{Z}$-linearly to all chains. When $v_0 \notin \{w_0, \dots, w_n\}$, the cone $[v_0, w_0, \dots, w_n]$ is a simplex of $K$ by the cone hypothesis, so $h_n$ takes values in $C_{n+1}(K)$. In negative degrees we set $h_{-1} = 0: C_{-1}(K) = 0 \to C_0(K)$.[/step]

[guided]The strategy for proving that $K$ has the homology of a point is to produce a chain homotopy between $\operatorname{id}_{C_\bullet(K)}$ and $i_\bullet \circ r_\bullet$. The latter map collapses every simplex to (a multiple of) $v_0$ in degree zero and vanishes in higher degrees. So we are looking for operators $h_n: C_n(K) \to C_{n+1}(K)$ satisfying \begin{align*} d_{n+1} \circ h_n + h_{n-1} \circ d_n &= \operatorname{id}_{C_n(K)} - i_n \circ r_n. \end{align*} What is the natural way to bridge a simplex $\sigma$ in $K$ with "the result of collapsing $\sigma$ onto $v_0$"? Geometrically, we form the cone from $\sigma$ up to $v_0$: this is a simplex one dimension higher whose boundary includes $\sigma$ on one face and a lower-dimensional structure on the rest. Algebraically, this is exactly the operation $\sigma \mapsto (v_0, \sigma)$. There is a subtlety: if $v_0$ already appears in $\sigma$, then the symbol $[v_0, w_0, \dots, w_n]$ would contain a repeated vertex and is not a simplex of $K$. In that case we must define $h_n(\sigma) = 0$. The case distinction is forced: there is no alternative consistent definition. The cone hypothesis ensures that in the remaining case — $v_0 \notin \sigma$ — the symbol $(v_0, \sigma)$ really is a simplex of $K$, so $h_n$ lands in $C_{n+1}(K)$ as required.[/guided]

[step:Verify the chain homotopy identity $d_{n+1} \circ h_n + h_{n-1} \circ d_n = \operatorname{id} - i_n \circ r_n$]It suffices to check the identity on the standard generators of $C_n(K)$: oriented $n$-simplices $\sigma = [w_0, \dots, w_n]$. We split into two cases according to whether $v_0$ is a vertex of $\sigma$. **Case 1: $v_0 \notin \{w_0, \dots, w_n\}$.** Then $h_n(\sigma) = [v_0, w_0, \dots, w_n]$, and by the definition of the simplicial boundary, \begin{align*} d_{n+1}(h_n(\sigma)) = d_{n+1}[v_0, w_0, \dots, w_n] = [w_0, \dots, w_n] - \sum_{i=0}^{n} (-1)^i [v_0, w_0, \dots, \widehat{w_i}, \dots, w_n]. \end{align*} On the other hand, $d_n \sigma = \sum_{i=0}^{n} (-1)^i [w_0, \dots, \widehat{w_i}, \dots, w_n]$. Each face $[w_0, \dots, \widehat{w_i}, \dots, w_n]$ also omits $v_0$, so by definition of $h$, \begin{align*} h_{n-1}(d_n \sigma) = \sum_{i=0}^{n} (-1)^i [v_0, w_0, \dots, \widehat{w_i}, \dots, w_n]. \end{align*} Adding, \begin{align*} (d_{n+1} h_n + h_{n-1} d_n)(\sigma) = [w_0, \dots, w_n] = \sigma. \end{align*} On this case, $r_n(\sigma) = 0$ for $n \geq 1$ (there are no simplices of dimension $\geq 1$ in $\{v_0\}$), and for $n = 0$ we have $r_0(\sigma) = v_0$ but $\sigma \neq v_0$; in either subcase $(i_n \circ r_n)(\sigma) = 0$ in $C_n(K)$ whenever $\sigma \neq v_0$. Hence $\sigma - (i_n \circ r_n)(\sigma) = \sigma$ in this case, matching the computation above. **Case 2: $v_0 \in \{w_0, \dots, w_n\}$.** By relabelling (and using that the oriented simplex is only defined up to a sign for permutations) write $\sigma = [v_0, w_1, \dots, w_n]$ so that $v_0$ is the first vertex. Then $h_n(\sigma) = 0$ by definition, so $d_{n+1}(h_n(\sigma)) = 0$. Now compute $d_n(\sigma)$: \begin{align*} d_n[v_0, w_1, \dots, w_n] = [w_1, \dots, w_n] - \sum_{i=1}^{n} (-1)^{i-1} [v_0, w_1, \dots, \widehat{w_i}, \dots, w_n]. \end{align*} Applying $h_{n-1}$: the first face $[w_1, \dots, w_n]$ does not contain $v_0$, so $h_{n-1}$ sends it to $[v_0, w_1, \dots, w_n] = \sigma$. Each remaining face $[v_0, w_1, \dots, \widehat{w_i}, \dots, w_n]$ does contain $v_0$, so $h_{n-1}$ sends it to $0$. Hence \begin{align*} h_{n-1}(d_n \sigma) = \sigma. \end{align*} Adding, \begin{align*} (d_{n+1} h_n + h_{n-1} d_n)(\sigma) = 0 + \sigma = \sigma. \end{align*} For the right-hand side: if $n \geq 1$, then $r_n(\sigma) = 0$ and $(i_n \circ r_n)(\sigma) = 0$, giving $\sigma - (i_n \circ r_n)(\sigma) = \sigma$. If $n = 0$, then $\sigma = v_0$ and $(i_0 \circ r_0)(v_0) = v_0 = \sigma$; but we also have to check the formula directly in degree $0$: $h_0(v_0) = 0$, $d_0 = 0$ (by convention), so the left-hand side is $0$, and the right-hand side is $v_0 - v_0 = 0$. Agreement. Combining the two cases, the chain homotopy identity \begin{align*} d_{n+1} \circ h_n + h_{n-1} \circ d_n = \operatorname{id}_{C_n(K)} - i_n \circ r_n \end{align*} holds for every $n \geq 0$.[/step]

[guided]The identity is the defining relation for a chain homotopy, and we verify it directly on generators. The two cases correspond to whether the coning construction sees the cone point or not. In Case 1 ($v_0 \notin \sigma$), coning genuinely produces a new $(n+1)$-simplex $(v_0, \sigma)$. Its boundary is built from $\sigma$ itself (the face obtained by deleting $v_0$) and the "side faces" $(v_0, \sigma^{(i)})$ where $\sigma^{(i)}$ denotes the $i$-th face of $\sigma$. The sign of the $\sigma$-face is $+1$ (since $v_0$ is the $0$-th vertex of $(v_0, \sigma)$, deleting it yields sign $(-1)^0 = 1$). The signs of the side faces $(v_0, \sigma^{(i)})$ are $-(-1)^i$: they come from deleting the $(i+1)$-th vertex of $(v_0, \sigma)$, which carries sign $(-1)^{i+1}$. The explicit sign tracking in the computation above shows that these side faces exactly cancel with the contribution of $h_{n-1}(d\sigma)$, which is a sum of cones on the $\sigma^{(i)}$ with signs $(-1)^i$. The cancellation is what allows only the "top face" $\sigma$ to survive, yielding the identity we want. In Case 2 ($v_0 \in \sigma$), the coning operation vanishes on $\sigma$ itself (since $h_n(\sigma) = 0$), so $d_{n+1} h_n \sigma = 0$. The only surviving contribution comes from $h_{n-1}(d_n \sigma)$. The boundary $d_n \sigma$ has one "special" face — the one obtained by deleting $v_0$ — which does not contain $v_0$ anymore, and all other faces still contain $v_0$. The latter are killed by $h_{n-1}$; the former, when coned, reproduces $\sigma$. This is how $\sigma$ reappears on the left-hand side with the right sign. The verification is a sign-tracking exercise, but the structure of the argument is: in either case, coning converts the boundary relation into exactly what we need to match $\operatorname{id} - i \circ r$.[/guided]

[step:Transfer the homology from the point to $K$] By the previous step, the chain maps $i_\bullet$ and $r_\bullet$ satisfy $r_\bullet \circ i_\bullet = \operatorname{id}$ (Step 1) and $i_\bullet \circ r_\bullet \simeq \operatorname{id}_{C_\bullet(K)}$ via the chain homotopy $h$ (Step 3). This is exactly the statement that $C_\bullet(\{v_0\})$ and $C_\bullet(K)$ are chain homotopy equivalent. By [Chain Homotopy Equivalences are Isomorphisms on Homology](/theorems/1925), the induced maps on homology \begin{align*} i_*: H_n(\{v_0\}) &\to H_n(K), & r_*: H_n(K) &\to H_n(\{v_0\}) \end{align*} are mutually inverse isomorphisms for every $n$. The homology of the one-point complex is computed directly: $C_0(\{v_0\}) = \mathbb{Z}$, $C_n(\{v_0\}) = 0$ for $n \neq 0$, and all boundary maps vanish. Hence \begin{align*} H_n(\{v_0\}) &= \begin{cases} \mathbb{Z} & n = 0, \\ 0 & n > 0. \end{cases} \end{align*} Transferring through the isomorphism, \begin{align*} H_n(K) &= \begin{cases} \mathbb{Z} & n = 0, \\ 0 & n > 0, \end{cases} \end{align*} which is the claim. This completes the proof. [/step]