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I am working on the first exercise in Hilton & Stammbach's "A Course in Homological Algebra", wherein I am given the following commutative diagram: enter image description here

I am asked to show that $\alpha'$ is an isomorphism with a diagram chase(I am relatively new to these).

I believe I have managed the injective part: if $\alpha'x'=0$, then $\mu'\alpha'x'=0$ as well, so that commutativity gives us that $\alpha\mu x'=0$. Yet $\alpha\mu$ is injective, so that $x'$ must be $0$. This implies injectivity of $\alpha'$.

Surjectivity is giving me a lot of trouble, however. I can chase $y'\in B'$ all over the place, but I always manage to get $\alpha\mu(x')=\mu'y'+$ extra term.

Any hints to offer on how to define $x'$?

user26857
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QTHalfTau
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I assume you're given the rows are exact! Use this! To illustrate how the argument begins, start with $b'\in B'$. Push it to $B$ by $\mu'$ and cover it by some $a\in A$. Commutativity and exactness at the lower row will give $\varepsilon \alpha a=0$, and exactness at the top row will give $\alpha a = \mu a'$. Continue to get the desired element.

Pedro
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  • Oooh, H&S mentioned in passing that a short exact sequence represents a module quotient, but I guess it didn't really sink in yet! – QTHalfTau May 22 '16 at 02:25
  • Oh, I have been using exactness to send elements leftward, but it should have been obvious that I can use it to determine that "two rights" make a 0 element. – QTHalfTau May 22 '16 at 02:30