Suppose $\phi, \psi: (X, x_0)\to(Y, y_0)$ are homotopic. $\exists\; H: X\times I\to Y:H(s,0)=\phi(s),\;H(s,1)=\psi(s)$. The claim is the induced homomorphisms $\phi_*, \psi_*$ on fundamental groups are equal according to this post. To prove this we have to show that $\phi\circ f$ is path homotopic to $\psi\circ f$ where $f$ is a loop at $x_0$. So we define $K:I^2\to Y$ as $K(s,t)=H(f(s),t)$ which ensures $K(s,0)=\phi\circ f(s)$ and $K(s, 1)=\psi\circ f(s)$ but to say that $K$ is a path homotopy we need $K(0,t),\; K(1, t)$ to be constant. How do we satisfy this requirement?
[Math] Doubt in proof : homotopic maps induce same homomorphism
algebraic-topologyfundamental-groupshomotopy-theory
Related Solutions
I will give an elementary proof of the problem using the fact, that $T^2$ is a topological group and that its universal cover is contractible. We will start with some constructions in homotopy theory of topological groups, which are required to understand the proof given below and added here for convenience.
First note, that $\mathbb R^2$ forms a topological group under addition: $(x,y) + (x',y') := (x+x',y+y')$ and that $\mathbb Z^2$ is a discrete normal subgroup thereof. We identify $T^2$ with the quotient of $\mathbb R^2$ by $\mathbb Z^2$, so that $T^2$ again becomes a topological group under addition. Moreover the quotient map $p: \mathbb R^2 \to T^2$ becomes a group homomorphism and is easily seen to be the universal covering projection ($\mathbb Z^2$ discrete subgroup $\Rightarrow$ $p$ is covering projection; $\mathbb R^2$ is contractible $\Rightarrow$ $p$ is universal).
Next, we observe, that for $[f],[g]\in [(X,\ast),(T^2,0)]$ the sum $[f]+[g] := [f+g]$ is well defined, turning $[(X,\ast),(T^2,0)]$ into a group. The same arguments show that $[(X,\ast),(\mathbb R^2,0)]$ is a group under point wise addition of representatives as well (as is $[(X,\ast),(G,1)]$ for any topological group $G$ with unit $1$), and that the map $p_\sharp : [(X,\ast),(\mathbb R^2,0)] \to [(X,\ast),(T^2,0)]$ given by $p_\sharp([f]) := [p \circ f]$ is a group homomorphism.
Now $\pi_1(T^2,0) = [(S^1,1), (T^2,0)]$ is a group in two ways, by means of composition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha \ast \beta]$ and by means of point wise addition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha + \beta]$.
Both operations share the same unit, the (class of the) constant loop sending everything to $0 \in T^2$ and denoted simply by $0: (S^1,1) \to (T^2,0)$. We can also observe, that for any loops $\alpha, \beta, \gamma, \delta$ we have $(\alpha + \beta) \ast (\gamma + \delta) = (\alpha + \gamma) \ast (\beta + \delta)$. Therefore $$[\alpha]+[\beta] = ([\alpha] \ast [0]) + ([0] \ast [\beta]) = ([\alpha] + [0]) \ast ([0] + [\beta]) = [\alpha] \ast [\beta],$$ hence the two operations are in fact the same on $\pi_1(T_2,0)$. The same argument can be used to show the analogous statement for $\pi_1(\mathbb R^2,0)$ (or $\pi_1(G,1)$ for any topological group $G$ with unit $1$).
Now back to the problem:
Given two maps $\varphi, \psi: T^2 \to T^2$, such that for some point $x \in T^2$, we have $\varphi(x) = \psi(x) = x$ and $\pi_1(\varphi) = \pi_1(\psi): \pi_1(T^2,x) \to \pi_1(T^2,x)$, we want to show $\varphi \simeq \psi$, where the homotopy can be taken relative to $x$. Replacing $\varphi$ with $\xi \mapsto \varphi(\xi + x) - x$ and $\psi$ with $\xi \mapsto \psi(\xi + x) - x$ if necessary, we may assume $x=0$. It will then suffice to show, that $\chi \simeq 0$, where $\chi := \varphi - \psi$.
Since the induced map $\pi_1(\chi): \pi_1(T^2,0) \to \pi_1(T^2,0)$ on fundamental groups is trivial (this is where we need all the constructions for topological groups), we can lift $\chi$ to a map $\bar{\chi}: (T^2,0) \to (\mathbb R^2,0)$ with $\chi = p \circ \bar{\chi}$.
We now define $H: T^2 \times I \to T^2$ by $H(x,t) = p(t\bar\chi(x))$, which is easily checked to be the required homotopy $0 \simeq \chi$.
No. Let $X = S^1$ (note that if there is a counterexample then composing with a loop that distinguishes the induced maps on fundamental groups shows that there is a counterexample where $X = S^1$) and let $Y$ be any space with nonabelian fundamental group. Pick a loop $h \in \pi_1(Y)$ such that there exists $g \in \pi_1(Y)$ with $ghg^{-1} \neq h$, and consider the free homotopy from $h$ to $ghg^{-1}$ given by transporting the loop around $g$.
Best Answer
In this context, when one says $\phi, \psi: (X, x_0)\to(Y, y_0)$ are homotopic, that means they are homotopic as maps $(X,x_0)\to(Y,y_0)$, not just as maps $X\to Y$. This means that every stage of the homotopy preserves the basepoint: $H(x_0,t)=y_0$ for all $t$. This guarantees that $K$ is a path homotopy.
(Without this stronger assumption on the homotopy from $\phi$ to $\psi$, it is not necessarily true that they induce the same map on fundamental groups. In general, they will differ by conjugation by the element of $\pi_1(Y,y_0)$ defined by the loop $t\mapsto H(x_0,t)$.)