Chapter 15: Problem 13
Prove that if \(\mathrm{A}\) is a strong deformation retract of \(\mathrm{X}\) then the inclusion map i: \(\mathrm{A} \rightarrow \mathrm{X}\) induces an isomorphism $$ \mathrm{i}_{*}: \pi(\mathrm{A}, \mathrm{a}) \rightarrow \pi(\mathrm{X}, \mathrm{a}) $$ for any point \(\mathrm{a} \in \mathrm{A}\).
Short Answer
Step by step solution
Understanding the Concept
Setup the Map i
Showing the Map i* Induces an Isomorphism
Verify Homotopy Equivalence
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Strong Deformation Retract
- \( F(x, 0) = x \) for all \( x \in \mathrm{X} \)
- \( F(x, 1) \in \mathrm{A} \) for all \( x \in \mathrm{X} \)
- \( F(a, t) = a \) for all \( a \in \mathrm{A}, \) and \( t \in [0, 1] \)
Fundamental Group
- The power of this group comes from its ability to distinguish between different topological spaces.
- If two spaces have different fundamental groups, they are not topologically equivalent.
- The operation to combine two loops, called the loop product, forms the group structure.
Homotopy
- \( H(x, 0) = f(x) \) and \( H(x, 1) = g(x) \)
- For each fixed \( t \), \( H(x, t) \) itself is a function from \( X \) to \( Y \)
Isomorphism
- An isomorphism is bijective, covering both injectiveness (one-to-one) and surjectiveness (onto).
- It preserves the group operation, meaning if two loops are combined in \( A \), their image in \( X \) under the inclusion map behaves identically.
- In essence, it shows that two groups are indistinguishably equivalent in structure and function, thus serving as an algebraic manifestation of topological similarity.