Corollary 2.4.23. Let \(f\colon X \to Y\) be a map of pointed animae with \(X\) and \(Y\) connected. Then \(f\) is an isomorphism if and only if the induced map \[ \Omega f\colon \Omega X \to \Omega Y \] is an isomorphism in \(\An \).
Proof. If \(f\) is an isomorphism, then clearly so is \(\Omega f\). Conversely, if \(\Omega f\) is an isomorphism, then it in particular induces isomorphisms on all homotopy groups, and using the identification \(\pi _n(X) \cong \pi _{n-1}(\Omega X)\) for \(n \geq 1\) it follows that the map \(\pi _n(f)\colon \pi _n(X) \to \pi _n(Y)\) is an isomorphism for all \(n \geq 1\). Since \(X\) and \(Y\) are connected, \(\pi _0(X) \to \pi _0(Y)\) is automatically a bijection, so \(f\) is an isomorphism by Proposition 2.4.22. □
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