This isn't homework, it's a proof left to the reader as I self study Munkre's 'Elements of Algebraic Topology'(adsbygoogle = window.adsbygoogle || []).push({});

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##

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# A Sequence induced by short exact sequence

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