Suppose we have a continuous map $f: X \to X$ such that $f_*: H_n(X) \to H_n(X)$ is the identity for all $n$. Is it true that in this case $f^*$ will be the identity on $H^n(X; G)$ for all $n$ and $G$? Does it hold when we additionally assume that $f$ is a homotopy equivalence?
Note that the first idea to simply use the naturality from the universal coefficient theorem does not work: Take for example the exact sequence
$$0 \to \mathbb{Z}_2 \to \mathbb{Z}_2 \oplus \mathbb{Z} \to \mathbb{Z} \to 0$$
if we choose $g: \mathbb{Z}_2 \oplus \mathbb{Z} \to \mathbb{Z}_2 \oplus \mathbb{Z}$ to send $(0, 1) \mapsto (1, 1)$, we can build a short commutative ladder with the identity on the left and right sides and $g$ in the middle. (The 5-lemma only implies that $g$ has to be an isomorphism.)
As a second, similar question suppose we have another map $g: Y \to Y$ that also induces the identity on homology. Will $(f \times g)_*$ be the identity on $H_n(X \times Y)$? Does the naturality from the Künneth formula help with this? (Of course similarly, what about cohomology?)
To provide a bit of context: I encountered this problem in reading Spanier's chapter on the Serre spectral sequence. The setting is as follows. We suppose that $p: E \to B$ is a fibration, $b_0 \in B$ and $F = p^{-1}(b_0)$. He calls $p$ orientable if for any loop $\gamma$ at $b_0$ in $B$ the associated homotopy equivalence of fibers $f: F \to F$ the induced map on homology $f_*$ is the identity. This is used to derive theorem 9.2.13. However, for the various analogues needed in section 9.4 the above statements seem necessary to me. I think there is no remark about using field coefficients or something like that.
Best Answer
Let $X=S^3\vee\Sigma\mathbb{RP}^2$ and consider the following map $f:X\to X$. On the $S^3$ summand, $f$ is the identity. On the $\Sigma\mathbb{RP}^2$ summand, $f$ is the composition of the pinch map $\Sigma\mathbb{RP}^2\to \Sigma\mathbb{RP}^2\vee \Sigma\mathbb{RP}^2$ with the map $\Sigma\mathbb{RP}^2\vee \Sigma\mathbb{RP}^2\to S^3\vee\Sigma\mathbb{RP}^2$ which is the identity on the second summand and on the first summand is the map $\Sigma\mathbb{RP}^2\to S^3$ that collapses the $2$-skeleton to a point.
To see that this $f$ induces the identity on homology, note that the only nontrivial reduced homology of $X$ is $H_2(X)\cong\mathbb{Z}/2$ via the $\Sigma\mathbb{RP}^2$ summand and $H_3(X)\cong\mathbb{Z}$ via the $S^3$ summand. So the induced map on $H_2$ can be computed by projecting down to just the $\Sigma\mathbb{RP}^2$ summand, and the induced map on $H_3$ can be computed by projecting down to just the $S^3$ summand, and $f$ becomes the identity in both cases.
However, $f$ does not induce the identity on cohomology. Indeed, $H^3(X;\mathbb{Z})\cong\mathbb{Z}\oplus\mathbb{Z}/2$ with the summands corresponding to the summands of $X$, and then $f^*$ maps $(1,0)$ to $(1,1)$ since the map $\Sigma\mathbb{RP}^2\to S^3$ used in the construction of $f$ maps $1\in H^3(S^3;\mathbb{Z})\cong\mathbb{Z}$ to $1\in H^3(\Sigma\mathbb{RP}^2;\mathbb{Z})$.
This is also an example of a map that induces the identity on homology with integer coefficients but not arbitrary coefficients, since the map $\Sigma\mathbb{RP}^2\to S^3$ is also nontrivial on homology with coefficients in $\mathbb{Z}/2$. Since there is a natural isomorphism $\tilde{H}_*(X\wedge\mathbb{RP}^2;\mathbb{Z})\cong \tilde{H}_{*-1}(X;\mathbb{Z}/2)$, this also gives a counterexample to your second question about products: $f\times 1:X\times\mathbb{RP}^2\to X\times\mathbb{RP}^2$ does not induce the identity on homology with integer coefficients.
(To provide some motivation, the starting point here is the observation that the obvious map of chain complexes from $0\to\mathbb{Z}\stackrel{2}{\to}\mathbb{Z}\to 0$ to $0\to\mathbb{Z}\to 0\to 0$ induces the $0$ map on homology but not on cohomology. Calling these chain complexes $Y$ and $Z$, you can then add this map to the identity map $Y\oplus Z\to Y\oplus Z$ to get a map $Y\oplus Z\to Y\oplus Z$ which is the identity on homology but not on cohomology. The topological example above is then just a way to realize this map of chain complexes with the homology of an actual space; the chain complexes $Y$ and $Z$ correspond to the spaces $\mathbb{RP}^2$ and $S^2$ but you have to suspend once in order to be able to add together maps on the level of spaces.)