[proofplan]
The induced-map construction is functorial — this is property (2) of [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939). We apply functoriality to the homeomorphism $f: |K| \to |L|$ and its continuous inverse $f^{-1}: |L| \to |K|$. Composing these in either order yields the identity continuous map, whose induced map on homology is the identity homomorphism (by functoriality applied to the identity). Hence $f_*$ and $(f^{-1})_*$ are two-sided inverses of each other, which forces $f_*$ to be an isomorphism.
[/proofplan]
[step:Verify that the identity continuous map induces the identity on homology]
Let $\operatorname{id}_{|K|}: |K| \to |K|$ denote the identity map, which is continuous. By [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939), it induces a well-defined homomorphism $(\operatorname{id}_{|K|})_*: H_n(K) \to H_n(K)$ for each $n$.
The identity simplicial map $\operatorname{id}_K: K \to K$ (sending each vertex to itself, hence each simplex to itself) is a simplicial approximation to $\operatorname{id}_{|K|}: |K| \to |K|$ at subdivision level $r = 0$: the star condition $\operatorname{id}_{|K|}(\operatorname{St}_K(v)) \subseteq \operatorname{St}_K(\operatorname{id}_K(v))$ holds trivially. By the independence of choice of approximation (property (1) of [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939)), we may compute $(\operatorname{id}_{|K|})_*$ using $\operatorname{id}_K$ at level $r = 0$:
\begin{align*}
(\operatorname{id}_{|K|})_* = (\operatorname{id}_K)_* \circ \nu_{K,0}^{-1}.
\end{align*}
At level $r = 0$, $\nu_{K,0} = \operatorname{id}_{H_n(K)}$ (the empty composition of subdivision isomorphisms), and $(\operatorname{id}_K)_*$ is the identity on $H_n(K)$ since the simplicial chain map induced by the identity simplicial map is the identity on each chain group. Therefore
\begin{align*}
(\operatorname{id}_{|K|})_* = \operatorname{id}_{H_n(K)}.
\end{align*}
The identical argument gives $(\operatorname{id}_{|L|})_* = \operatorname{id}_{H_n(L)}$.
[/step]
[step:Apply functoriality to $f \circ f^{-1}$ and $f^{-1} \circ f$]
Since $f: |K| \to |L|$ is a homeomorphism, its inverse $f^{-1}: |L| \to |K|$ is a continuous map. Apply property (2) of [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939) — functoriality — to the continuous pair $(f^{-1}, f)$:
\begin{align*}
(f^{-1} \circ f)_* = (f^{-1})_* \circ f_* \quad \text{as maps} \quad H_n(K) \to H_n(K).
\end{align*}
But $f^{-1} \circ f = \operatorname{id}_{|K|}$ as continuous maps $|K| \to |K|$, so by Step 1,
\begin{align*}
(f^{-1})_* \circ f_* = (\operatorname{id}_{|K|})_* = \operatorname{id}_{H_n(K)}.
\end{align*}
Symmetrically, applying functoriality to $(f, f^{-1})$ and using $f \circ f^{-1} = \operatorname{id}_{|L|}$:
\begin{align*}
f_* \circ (f^{-1})_* = (\operatorname{id}_{|L|})_* = \operatorname{id}_{H_n(L)}.
\end{align*}
[/step]
[step:Conclude that $f_*$ is an isomorphism with inverse $(f^{-1})_*$]
The previous step exhibits $(f^{-1})_*: H_n(L) \to H_n(K)$ as a two-sided inverse to $f_*: H_n(K) \to H_n(L)$. A homomorphism with a two-sided inverse is an isomorphism, so $f_*$ is an isomorphism of abelian groups for every $n$. This holds for all $n$, completing the proof.
[/step]
