How to prove the inclusion is a homotopy equivalence?

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How to prove the inclusion is a homotopy equivalence?

Homework Statement



A deformation retraction in the weak sense of a space X to a subspace A is a homotopy f_t: X\rightarrow X such that f_0=Id_x, f_1(X)\subset A, and f_t(A)\subset A for all t. Show that if X deformation retracts to A in this weak sense, then the inclusion A\hookrightarrow X is a homotopy equivalence.

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The Attempt at a Solution



I have one problem. Let r: X\rightarrow A be the retraction. I can prove that ri\simeq Id_A. Then I have to prove that ir\simeq Id_X. Then there is a problem. To achieve this goal, I have to use the homotopy f_t. This homotopy is defined on the domain X. It seems that I have to compose f_t with inclusion, but the codomain of f_t is different from the domain of i. So I do not know how to do. Please give some hints. Thanks a lot!:shy:
 
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No one knows the answer? What a pity! Hope someone can help me!
 


qinglong.1397 said:
It seems that I have to compose f_t with inclusion, but the codomain of f_t is different from the domain of i.

Isn't it more important that i be defined on the image of f_t?
 


fzero said:
Isn't it more important that i be defined on the image of f_t?

Can i be defined on the image of f_t?

I do not think so. That is because the domain of i and the codomain of f_t are different sets. For example, if t=0, f_0=Id_X whose codomain is X, but the domain of i is A. So the composition if_0 makes none sense.
 


Is there someone who can help me?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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