# Math Help - Commutative Diagram, Modules

1. ## Commutative Diagram, Modules

Suppose

is a commutative diagram of modules over some ring $R$. Suppose the rows are exact and $g$ and $h$ are bijective. Prove $f$ is bijective.

Attempt:
I know the snake lemma and five lemma. This looks more like a diagram chasing question. In the five lemma and snake lemma I can use the commutativity of the diagram, but here, $f$ is on the left, so I don't know how to use the commutativity of the diagram. Any hints for showing this would be nice. Thanks.

2. Originally Posted by xianghu21
Suppose

is a commutative diagram of modules over some ring $R$. Suppose the rows are exact and $g$ and $h$ are bijective. Prove $f$ is bijective.

Attempt:
I know the snake lemma and five lemma. This looks more like a diagram chasing question. In the five lemma and snake lemma I can use the commutativity of the diagram, but here, $f$ is on the left, so I don't know how to use the commutativity of the diagram. Any hints for showing this would be nice. Thanks.
try to solve problems like this (although they're very easy) without using theorems and lemmas! it's good for better learning:

call the maps in the first row $\alpha, \beta$ and in the second row $\alpha', \beta'.$ suppose $f(a)=0.$ then $0=\alpha'f(a)=g \alpha(a)$ and thus $a=0$ because both $g, \alpha$ are injective. so $f$ is injective.

now let $a' \in A'.$ then $g(b)=\alpha'(a'),$ for some $b \in B,$ because $g$ is surjective. so $0=\beta' \alpha'(a')=\beta'g(b)=h \beta(b).$ thus $\beta(b)=0,$ because $h$ is injective. so $b \in \text{Im}(\alpha).$ hence

$b=\alpha(a),$ for some $a \in A$ and therefore $\alpha' f(a)=g \alpha(a)=g(b)=\alpha'(a').$ thus $f(a)=a'$ proving that $f$ is surjective.

Remark: you might have noticed that we didn't use surjectivity of $h.$ the point here is that " $h$ is injective $\Longleftrightarrow h$ is bijective". i'll leave this, as an easy exercise, for you to prove.