- #1
PsychonautQQ
- 784
- 10
This isn't homework, it's a proof left to the reader as I self study Munkre's 'Elements of Algebraic Topology'
Prove that if the sequence
##A_1 --> A_2 --> A_3 --> A_4 --> A_5## is exact
Then so is the induced sequence:
##0 --> cok(a_1) --> A_3 --> ker(a_4) --> 0##
where ##a_1## and ##a_4## are the maps from ##A_1 --> A_2## and ##A_4 --> A_5## respectively.
Also, let ##b_i## represent the map from the i'th group in the induced sequence to the (i+1)'th
First of all, what exactly is an induced sequence? Secondly, I started trying to figure out the proof and failed. Here are Some of my thoughts:
If ##x,y \in cok(a_1)## and ##b_2(x)=b_2(y)##, wwts ##x=y##, because that will mean that ##b_2## is injective, which it must be if it's going to be exact because ##im(b_1) = 0 = ker(b_2)##.
Since ##x,y \in cok(a_1)##, there are no elements of ##A_1## that map to them in ##A_2##.
And ahhh yeah actually I'm just confused... anyone got any tips for this proof?
This means there are no elements of ##A_1## that map to ##x,y## in ##A_2##
Prove that if the sequence
##A_1 --> A_2 --> A_3 --> A_4 --> A_5## is exact
Then so is the induced sequence:
##0 --> cok(a_1) --> A_3 --> ker(a_4) --> 0##
where ##a_1## and ##a_4## are the maps from ##A_1 --> A_2## and ##A_4 --> A_5## respectively.
Also, let ##b_i## represent the map from the i'th group in the induced sequence to the (i+1)'th
First of all, what exactly is an induced sequence? Secondly, I started trying to figure out the proof and failed. Here are Some of my thoughts:
If ##x,y \in cok(a_1)## and ##b_2(x)=b_2(y)##, wwts ##x=y##, because that will mean that ##b_2## is injective, which it must be if it's going to be exact because ##im(b_1) = 0 = ker(b_2)##.
Since ##x,y \in cok(a_1)##, there are no elements of ##A_1## that map to them in ##A_2##.
And ahhh yeah actually I'm just confused... anyone got any tips for this proof?
This means there are no elements of ##A_1## that map to ##x,y## in ##A_2##