Let $A$ be a retract of $X$. Show that if $i:A\to X$ is the inclusion operator , then $i_*:H_n(A) \to H_n(X)$ is an isomorphism
(It's easy to see that it's a monomorphism. But why is it an epimorphism?)
Thanks!
Let $A$ be a retract of $X$. Show that if $i:A\to X$ is the inclusion operator , then $i_*:H_n(A) \to H_n(X)$ is an isomorphism
(It's easy to see that it's a monomorphism. But why is it an epimorphism?)
Thanks!
As others have mentioned, this statement is not true without a stronger condition on $r$: that it is a deformation retraction. That is, you also want $i\circ r$ to be homotopic to the identity on $X$. Then $(i\circ r)_*=(\text{id}_X)_*$ is the identity so $i_*$ is surjective. Combine this with the injectivity of $i_*$ to get that it is an isomorphism.
Let $i:A \to X$ be the inclusion and $r:X \to A$ the retraction. By definition $r \circ i = \text{id}_A$. Then use the fact homology is a functor.