A weak equivalence has vanishing mapping-cylinder relative groups

Both endpoint inclusions are closed embeddings, including for non-Hausdorff spaces. They are continuous injective maps. For a closed $C\subseteq X$, the inverse image of $j(C)$ under the quotient map is just $C\times \{1\}$, closed in the disjoint union. Thus $j(C)$ is closed in $M_f$, which proves that $j$ is a closed embedding. For closed $E\subseteq Y$, the inverse image of $k(E)$ is $E\bigsqcup (f^{-1}(E)\times \{0\})$, also closed. Thus $k$ is a closed embedding. In particular the subspace pair in the statement really uses the given topology of $X$.

Define $r:M_f\to Y$ by $r(k(y))=y$ and $r([x,s])=f(x)$. The defining maps on the disjoint summands are continuous and respect the identifications, so [F6] makes $r$ continuous, with $rj=f$ and $rk={\mathrm{i}\mathrm{d}}_Y$. The formulas $D(k(y),t)=k(y),D([x,s],t)=[x,(1-t)s]$ agree at the gluing end and descend continuously by [F3]. They define a homotopy from ${\mathrm{i}\mathrm{d}}_{M_f}$ to $kr$, fixing $k(Y)$ pointwise. Each $z\in M_f$ is joined by its track to $k(r(z))$, so ${\pi }_0(k){\pi }_0(r)$ is the identity; the other composite is the identity because $rk$ is. Therefore ${\pi }_0(r)$ is bijective.

Fix $x\in X$ and put $y=f(x)$. At the basepoint $k(y)$ the homotopy $D$ is based. Thus [F4] and $rk=\mathrm{i}\mathrm{d}$ show that $k_{\ast }:{\pi }_n(Y,y)⟶{\pi }_n(M_f,k(y))$ is an isomorphism with inverse induced by $r$. At the source endpoint $j(x)$, the track is ${\gamma }_x(t)=[x,1-t]$, running from $j(x)$ to $k(y)$. Apply [F5] to $D$, now with domain based at $j(x)$: ${\mathrm{i}\mathrm{d}}_{{\pi }_n(M_f,j(x))}={\beta }_{{\gamma }_x}{k}_{\ast }{r}_{\ast }\mathrm{.}$ Both ${\beta }_{{\gamma }_x}$ and the displayed $k_{\ast }$ are isomorphisms. Consequently $r_{\ast }:{\pi }_n(M_f,j(x))\to {\pi }_n(Y,y)$ is their inverse composite, and is an isomorphism for every $n\ge 1$. This uses the actual track; it does not mistake $k$ for a based inverse at $j(x)$.

Since $f=rj$, steps 1.2 and 2.1 imply that $f$ is weak precisely when $j$ is bijective on components and induces isomorphisms on all positive groups at each $x\in X$. Suppose first these conditions hold. For $n\ge 2$ and $\alpha \in {\pi }_n(M_f,X,j(x))$, its boundary belongs to the kernel of ${\pi }_{n-1}(X,x)\to {\pi }_{n-1}(M_f,j(x))$. That kernel is trivial, so exactness [F2] puts $\alpha$ in the image of ${\pi }_n(M_f,j(x))$. Surjectivity from ${\pi }_n(X,x)$ then makes this image trivial by exactness at ${\pi }_n(M_f,j(x))$. Thus $\alpha$ is the distinguished element. This reasoning also works for $n=2$, without assuming that the relative group is abelian.

Conversely suppose the component condition and all the relative trivialities in the statement. Fix $x\in X$ and $n\ge 1$. In the exact segment ${\pi }_{n+1}(M_f,X,j(x))⟶{\pi }_n(X,x)\stackrel{j_{\ast }}{\to }{\pi }_n(M_f,j(x))⟶{\pi }_n(M_f,X,j(x)),$ the left and right relative terms are trivial. Exactness at ${\pi }_n(X,x)$ gives a trivial kernel, so its homomorphism $j_{\ast }$ is injective. Exactness at ${\pi }_n(M_f,j(x))$ gives surjectivity, since the next map takes everything to the distinguished element. This includes $n=1$, whose rightmost term is only a pointed set. Hence $j_{\ast }$ is an isomorphism in every positive degree. Combining with the separately assumed component bijection and steps 1.2 and 2.1 proves that $f$ is weak.

In relative degree one let $\alpha$ be any path class from a point of $X$ to $j(x)$. Its boundary is a component of $X$ mapping to the component of $j(x)$. Injectivity of ${\pi }_0(j)$ implies that this boundary is the distinguished component of $x$. Exactness of the pointed tail [F2] puts $\alpha$ in the image of ${\pi }_1(M_f,j(x))$. Surjectivity of ${\pi }_1(X,x)\to {\pi }_1(M_f,j(x))$ and exactness at the latter group show that its whole image in the relative pointed set is the distinguished point. Therefore ${\pi }_1(M_f,X,j(x))$ has one element. This argument uses no subtraction or group operation on that pointed set.

If $X$ is empty, $M_f=Y$ and relative basepoint assertions are vacuous, but component bijectivity on either side forces $Y$ empty. Thus the equivalence still holds. Equal endpoint images or a constant map cause no problem in the quotient formulas: the free end remains embedded, and the deformation fixes every point of the included target. The homotopy has exactly the stated values at $t=0,1$ and each source basepoint uses its explicitly prescribed track. All arguments are formulas, exactness or an argument at one arbitrary point/class. No selection of representatives or choice principle is used. Steps 3.1 and 4.1 prove the forward direction, and step 3.2 proves the converse.