Reading mathematical claims in an eng book after reading math books makes you want to cry

@EE18 Who's this "you", white man?

me myself and i

xander: oh, and the kids are going to get that reference. OK.

@leslietownes Okay, so we're both old.

can anyone please explain where the contradiction is?

god, what a convoluted proof

anyway, the contradiction comes from writing down the induced maps on cohomology rings

In past, I have done it using fibrations.

think about what $f$ does on the generator

But this time fibration sequences are not allowed [I didn't impose this condition].

anyways, I was thinking: f o h= g so passing onto cohomology: h* o f*= g* and on S^{2n+1}, g* being nullhomotopic so 0 so h*o f*=0 on S^{2n+1}.

this gives j* o pi* o f* =0 whence j* o (f o pi)*=0

But since f o j= id, we have j* o f*= id

this contradicts the earlier equality.

But there must be something wrong in it. :(

the equality $f\circ h=g$ doesn't imply anything meaningful at the level of homotopy invariants, because $D^{2n+2}$ is contractible

you can derive a contradiction from $f\circ j=\mathrm{id}$ and that alone