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This question is on an old qualifying exam for my institution:

Let $Y$ be a space and $f:Y\to Y$ a self-mapping. Let $X$ be the mapping torus of $f$ (i.e. the space obtained from $Y\times I$ by identifying $(y,1)$ and $(f(y),0)$ for $y\in Y$). Prove that $H_1(X;\Bbb Z)\cong H_1(Y;\Bbb Z)/\mathrm{im}(\mathrm{id}-f_*)$ where $f_*:H_1(Y;\Bbb Z)\to H_1(Y;\Bbb Z)$ is the induced map.

Now, the reason I am confused is because if you take $f$ to be the identity, then $X$ is just $Y\times S^1$ and in particular if we take $Y$ to be, say, a point then $X=S^1$ and we get

$$\Bbb Z=H_1(X;\Bbb Z)=H_1(Y;\Bbb Z)/\mathrm{im}(\mathrm{id}-f_*)=0.$$

Is there some flaw in my logic or is the question missing some assumption?

Alex Mathers
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    This see incorrect to me without further hypothesis. There is a long exact sequence $H_n(X) \to H_n(X) \to H_n(T_f) \to H_{n-1}(X) \to \dots$ where the first map is $id-f_*$. It is obtained by taking the complex $$H_n(X \times I,X \times \partial I) \to H_{n-1}(X \times \partial I) \to H_{n-1}(X \times I) \to \dots$$ and studying the maps induced by the quotient $q:(X \times I, \times \partial I) \to T_f$ and its restrictions – Andres Mejia Aug 22 '18 at 00:42
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    I just found a reference. See hatcher 2.48. – Andres Mejia Aug 22 '18 at 00:45
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    @AndresMejia Dope, just wanted a sanity check. Thanks Andres – Alex Mathers Aug 22 '18 at 05:36
  • Write $X$ as the pushout of $Y\times I\xleftarrow{(i_0,i_1)} Y\vee Y\xrightarrow{(1,f)}Y$ and use Meyer-Vietoris. You get a short exact sequence which looks something like $H_1Y\oplus\ H_1Y\xrightarrow{(1-f_*\oplus 1)\oplus(0,1)} H_1Y\oplus H_1 Y\rightarrow H_1X\rightarrow 0$ after some fiddling. You can cancel a copy of $H_1Y$ to identify $H_1X$ with the quotient you want. Note that when $f=id_Y$, the homotopy type of $X_{id}$ is the same as the homotopy pushout of $Y\xleftarrow{\nabla}Y\vee V\xrightarrow{\nabla} Y$, which is known to be weakly contractible. – Tyrone Aug 22 '18 at 12:04
  • ... when $\pi_1Y$ contains no perfect subgroups. – Tyrone Aug 22 '18 at 15:02
  • @AlexMathers no problem! tyrone, it's the homotopy pushout given that $\pi_1(Y)$ has no perfect subgroups or it is weakly contractible in that case? – Andres Mejia Aug 22 '18 at 18:11

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