operationsres
- 99
- 0
Preface: Can't understand notation from 1st year Real Analysis textbook that they pulled out of thin air without defining.
1. The thing statement that they're proving
Let A and B be sets. Moreover, let [itex]f : A \rightarrow B[/itex] be some injective map. Calculate [itex]f^{-1} \circ f[/itex].
SOLUTION: To every [itex]y \in f(A)[/itex], the map [itex]f^{-1}[/itex] associates the corresponding [itex]x \in A[/itex] which satisfies [itex]f(x) = y[/itex]. In particular, it associates with [itex]f(x)[/itex] the element [itex]x[/itex] for all [itex]x \in A[/itex]. Hence:
[itex]f^{-1} \circ f = id_A[/itex], [itex]f \circ f^{-1} = id_{f(A)}[/itex],
where for every set [itex]C[/itex] the corresponding map [itex]id_C : C \rightarrow C[/itex] is defined by:
[itex]id_C (x) := C[/itex]
for all [itex]x \in C[/itex].
2. My confusion
What's [itex]id[/itex]?
1. The thing statement that they're proving
Let A and B be sets. Moreover, let [itex]f : A \rightarrow B[/itex] be some injective map. Calculate [itex]f^{-1} \circ f[/itex].
SOLUTION: To every [itex]y \in f(A)[/itex], the map [itex]f^{-1}[/itex] associates the corresponding [itex]x \in A[/itex] which satisfies [itex]f(x) = y[/itex]. In particular, it associates with [itex]f(x)[/itex] the element [itex]x[/itex] for all [itex]x \in A[/itex]. Hence:
[itex]f^{-1} \circ f = id_A[/itex], [itex]f \circ f^{-1} = id_{f(A)}[/itex],
where for every set [itex]C[/itex] the corresponding map [itex]id_C : C \rightarrow C[/itex] is defined by:
[itex]id_C (x) := C[/itex]
for all [itex]x \in C[/itex].
2. My confusion
What's [itex]id[/itex]?