- #1
roam
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Homework Statement
Let A,B be sets and let [tex]f:A \rightarrow B[/tex] and [tex]g,h:B \rightarrow A[/tex] be functions.
(1) Suppose [tex]h o f[/tex] is an injective map from [tex]A[/tex] to itself. Show that [tex]f[/tex] is injective.
(2) Suppose now that [tex]f o g = 1_{B}[/tex] and [tex]hof = 1_{A}[/tex]. Show that [tex]f[/tex] is bijective and [tex]g=h[/tex].
P.S. given that f is surjective.
Homework Equations
The Attempt at a Solution
(1) [tex]h o f[/tex] is 1-1 [tex]\Leftrightarrow \forall a,a' \in A[/tex] such that [tex]a \neq a'[/tex], [tex]h o f(a) \neq h o f(a')[/tex]
[tex]\Leftrightarrow h(f(a)) \neq h(f(a'))[/tex]
[tex]\Leftrightarrow f(a) \neq f(a')[/tex]
I'm fine with part (1)
(2) I need some help to write a proper proof for this one.
[tex]f[/tex] is an injective map from my previus work in part (1), I also think [tex]h \circ f = 1_A[/tex] but I don't know how to justify this. ([tex]1_A[/tex] & [tex]1_b[/tex] notation represents the identity).
[tex]f[/tex] is surjective, if it is not, then [tex]f o g[/tex] also isn't surjective, but [tex]f \circ g = 1_B[/tex] is surjective.
Further, we have [tex]g= 1_A \circ g=h \circ f \circ g=h \circ 1_B=h[/tex], since composition of functions is associative.
Any help is appreciated! This is VERY urgent!