# Composition of Maps

1. Aug 26, 2009

### roam

1. The problem statement, all variables and given/known data
Let A,B be sets and let $$f:A \rightarrow B$$ and $$g,h:B \rightarrow A$$ be functions.

(1) Suppose $$h o f$$ is an injective map from $$A$$ to itself. Show that $$f$$ is injective.

(2) Suppose now that $$f o g = 1_{B}$$ and $$hof = 1_{A}$$. Show that $$f$$ is bijective and $$g=h$$.

P.S. given that f is surjective.

2. Relevant equations

3. The attempt at a solution

(1) $$h o f$$ is 1-1 $$\Leftrightarrow \forall a,a' \in A$$ such that $$a \neq a'$$, $$h o f(a) \neq h o f(a')$$

$$\Leftrightarrow h(f(a)) \neq h(f(a'))$$

$$\Leftrightarrow f(a) \neq f(a')$$

I'm fine with part (1)

(2) I need some help to write a proper proof for this one.

$$f$$ is an injective map from my previus work in part (1), I also think $$h \circ f = 1_A$$ but I don't know how to justify this. ($$1_A$$ & $$1_b$$ notation represents the identity).

$$f$$ is surjective, if it is not, then $$f o g$$ also isn't surjective, but $$f \circ g = 1_B$$ is surjective.
Further, we have $$g= 1_A \circ g=h \circ f \circ g=h \circ 1_B=h$$, since composition of functions is associative.

Any help is appreciated! This is VERY urgent!

2. Aug 26, 2009

### HallsofIvy

You say you are given that $$h \circ f= 1_A$$, you don't need to prove it!

Again, you say you are given that f is surjective!

3. Aug 26, 2009

### roam

But I still need to prove this:

I don't know what to write down as the "proof"...

4. Aug 26, 2009

### Elucidus

f is bijective if it is injective and surjective (which I believe you showed). The proof of g = h you already did (owing to associativity of composition).

In your proof of part (1), I'd be careful with your implication arrows. They all go to the right just fine, only some go to the left. There is no need to explore the leftward implications. You only need to show $a \neq a' \Rightarrow f(a) \neq f(a')$.

--Elucidus