Sina

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seen Aug 3 '12 at 7:04

Aug
3
comment Hatcher Question 0.27
Yes and when we deformation retract $M_f$ to A (or to say A$\times$1) then arent we left with $X\times0 \cup A\times I$?
Aug
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awarded  Scholar
Aug
3
accepted Hatcher Algebraic Topology 0.24
Aug
1
asked Hatcher Algebraic Topology 0.24
Aug
1
comment Hatcher Question 0.27
On the first line shouldnt you say something like $X\times \{0\}\cup A\times I\cup M_{f} \cong X\times \{0\}\cup A\times I$ since $M_f \cong A$. Where does 2I come from?
Aug
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awarded  Student
Aug
1
asked Hatcher Question 0.27