Singular Cochains Mayer Vietoris and Smooth Singular Comparison — Examples

The image of $\kappa$ is $\{0\}\subset U$, so it is already small. The image $[0,1]$ of $\sigma$ is in neither $U$ nor $V$, since $1\mathrm{\notin }U$ and $0\mathrm{\notin }V$. Thus its least depth is positive. Both paths extend smoothly to all real parameters.

The cone convention [F2] gives $S\sigma =\alpha -\beta$, where $\alpha (t)=(1+t)\mathrm{/}2$ and $\beta (t)=(1-t)\mathrm{/}2$. Their images are respectively $[1\mathrm{/}2,1]\subset V$ and $[0,1\mathrm{/}2]\subset U$. Thus $S\sigma$ is small and the least depth of $\sigma$ is exactly one. Since both halves of the constant path are the same constant path, $S\kappa =\kappa -\kappa =0$. Consequently $Sc=\alpha -\beta$ is small.

This exhibits different least depths within a finite chain, while the common bound one works for the entire chain. The two terms of $c$ are distinct basis maps, so the nonsmall $\sigma$ does not cancel before subdivision. Zero coefficients or the empty chain would have no such obligation. Endpoint inclusions above are strict relative to the cover thresholds, and the degenerate constant path has been computed rather than discarded. No choice is used.

We have $U\cap V=(-1,1)$, so $p$ lies in the overlap and $q$ lies in $U\setminus V$. By [F1], $(E_U\eta )([p])=2$ and $(E_U\eta )([q])=0$. Linearity therefore gives $(E_U\eta )(3[p]-[q])=3\cdot 2-0=6$.

Restriction back to the overlap equals $\eta$ on every vertex: it has value two at $p$ and zero elsewhere, hence agrees on every finite chain. More generally, for any specified overlap cochain and finite chain $\sum a_s[s]$ in $U$, the value is exactly ${\sum }_{s\text{ in the overlap}}{a}_s\eta (s)$. An empty retained index set gives zero and one retained term gives its coefficient times its value. This is a degreewise extension, not an assertion that extension commutes with coboundary. No basis or representatives are chosen.

The affine map $L:A^k\to {\mathbb{R}}^n$, $L(\lambda )={\sum }_i{\lambda }_i{v}_i$, is smooth. Convexity implies $L({\mathrm{\Delta }}^k)\subset B$. Therefore $O=L^{-1}(B)$ is open in $A^k$ and contains the entire simplex. The smooth map ${\psi }^{-1}\circ L:O\to W\subset M$ is the extension required by [F1]. This constructs one common neighbourhood directly.

To make the uniform margin explicit, $K=L({\mathrm{\Delta }}^k)$ is compact. If $B=B(b,R)$, the continuous function $y\mapsto \mathrm{\parallel }y-b\mathrm{\parallel }$ attains a maximum $r<R$ on $K$. Put $\epsilon =(R-r)\mathrm{/}2>0$. Any point within distance $\epsilon$ of $K$ lies in $B$ by the triangle inequality. Hence $L^{-1}(\{y:\mathrm{dist}(y,K)<\epsilon \})$ is a single open neighbourhood of the closed simplex on which the same extension is defined. For example, with $B=(-2,2)$ and vertices $-1,1$, the path is $\sigma (t)=2t-1$, whose extension remains in $B$ for $-1\mathrm{/}2<t<3\mathrm{/}2$.

For $k=0$ this gives the constant extension on the one-point affine space. For repeated or coincident vertices the affine formula is still smooth and may be constant; no independence is needed. The construction includes all faces and endpoints because $O$ contains the closed simplex. Empty balls cannot carry the given vertex tuple; in dimension zero the ball is a point and the same formula is constant. Only a supplied finite tuple is used, so no AC is required.

Each term is nonnegative and at most $2^{-n-1}$ by [F3]. Thus the sum is well-defined and $0\le T\le 1$. By [F2] it is continuous, so it is a singular one-simplex. Direct substitution gives $T(0)=T(1)=0$ and $T(1\mathrm{/}2)=1\mathrm{/}2$, because every term with $n\ge 1$ vanishes there. The path is therefore nonconstant despite its equal endpoints.

To spell out the differentiability obstruction supplied by [F2], take the nested adjacent dyadic interval $[u_N,v_N]$ of length $2^{-N}$ containing the parameter, using the interval to the right at a dyadic point and the interval to the left at $1$. All summands of index at least $N$ vanish at its endpoints. Each earlier summand is affine there with slope ${\epsilon }_n\in \{-1,1\}$. The secant slope of $T$ is ${\sum }_{n<N}{\epsilon }_n$. Nested intervals preserve the earlier slopes, so consecutive secant slopes differ by one in absolute value and cannot converge to a finite value. If a finite derivative existed, the two endpoint quotients would tend to it, and their convex combination, this secant slope, would also tend to it. At a dyadic point or endpoint the appropriate one-sided quotient gives the same contradiction. This verifies exactly the finite-derivative assertion needed here.

A smooth extension in [F1] would give a finite derivative at every interior parameter, contradicting step 2.1. Thus this example meets the stronger nowhere-differentiable requirement, not only failure at one cusp. There is no empty-domain case for a singular one-simplex. A point target would yield a constant smooth map and is not this witness. No infinite selections are used: the series and dyadic intervals are specified arithmetically.

Regard $[0,1]$ as ${\mathrm{\Delta }}^1$. Its two faces are the separate points $0$ and $1$. Prescribe their smooth zero-simplices with values $p,q$ and the constant homotopies $H_0(0,s)=p$, $H_1(1,s)=q$. The faces have empty intersection, so compatibility is vacuous. Under [F2], [F1] supplies a strict smooth path $g$ and a continuous homotopy $H$ satisfying $H(0,s)=p$ and $H(1,s)=q$ for every $s$, as required.

The neighbourhood hypothesis behind [F1] is concrete in this dimension: disjoint small affine neighbourhoods of the two endpoint faces carry the constant smooth maps $p$ and $q$. The relative-smoothing proof first changes the continuous path, keeping the endpoints fixed, to agree with such a smooth neighbourhood extension near the endpoint union; only then does it apply relative Whitney approximation. Thus no claim is made that mere equality of endpoint values already means smoothness near those endpoints.

In the explicit real example define $H(t,s)=(1-s)\mathrm{\mid }t-1\mathrm{/}2\mathrm{\mid }+s\mathrm{/}2$. This is continuous, equals $f$ at $s=0$, equals the constant smooth path $g$ at $s=1$, and has $H(0,s)=H(1,s)=(1-s)\mathrm{/}2+s\mathrm{/}2=1\mathrm{/}2$ at every time. Its middle value is $H(1\mathrm{/}2,s)=s\mathrm{/}2$, exhibiting the actual change of the path. If the original path is smooth, [F1] also permits the constant homotopy with $g=f$, including constant paths and a one-point target. An empty target admits no path. The general assertion inherits only countable choice; the displayed real formula needs none.

There is exactly one simplex $s_k:{\mathrm{\Delta }}^k\to P$ in each nonnegative degree, so $C_k(P;\mathbb{R})=\mathbb{R}[s_k]$ and its real dual is $\mathbb{R}$ by evaluation on $s_k$. For $k>0$ all faces equal $s_{k-1}$, hence $\mathrm{\partial }{s}_k=({\sum }_{i=0}^k(-1{)}^i)s_{k-1}$. Pairing consecutive signs gives scalar zero for odd $k$ and one for even $k$. The degree-zero boundary is zero by convention.

Thus ${\delta }^k$ is zero for even $k\ge 0$ and identity for odd $k$. In degree zero, kernel is $\mathbb{R}$ and the image from degree minus one is zero. In positive even degree, kernel is $\mathbb{R}$ and the previous differential is identity, so the quotient is zero. In odd degree, kernel is zero and the previous image is zero, again giving zero. Negative cochain groups and cohomology are zero. This proves all claimed values by the actual quotient definition.

All higher point simplices are degenerate but were retained; discarding them without changing complexes would not be the calculation above. In particular the sole edge has boundary $[P]-[P]=0$ and the sole triangle has boundary $s_1-s_1+s_1=s_1$. The zero vector is the unique class in every positive group. For comparison the empty space has no basis simplices, hence zero in all cochain and cohomology degrees. These canonical identifications use no choice.

By [F1], ${\mathrm{\partial }}_n{\mathrm{\partial }}_{n+1}=0$, and hence for every cochain $f$ one has ${\delta }^{n+1}{\delta }^{n}f=f{\mathrm{\partial }}_{n+1}{\mathrm{\partial }}_{n+2}=0$. Thus the image in [F2] is a vector subspace of the kernel and the quotient exists. Both kernel and image are specified sets, and forming their quotient makes no selection of representatives. This verifies the choice-free definition.

If $f$ is a cocycle and $z$ a cycle, set $\epsilon ([f])([z])=f(z)$. Replacing $z$ by $z+\mathrm{\partial }c$ changes this value by $f(\mathrm{\partial }c)=(\delta f)(c)=0$. Replacing $f$ by $f+\delta g$ changes it by $g(\mathrm{\partial }z)=0$. Addition and scalar multiplication commute with evaluation, so it defines the claimed linear map on the two quotients. For a continuous map $u:X\to Y$, postcomposition on simplices commutes with each face, and hence with the signed boundary. Its chain map $u_{\mathrm{\#}}$ therefore satisfies $f(u_{\mathrm{\#}}z)=(f\circ u_{\mathrm{\#}})(z)$, which is the naturality identity on classes. Neither construction uses AC.

Assume AC for this step. Apply the positive branch of [F3] to the real complex [F1]. Concretely, any functional on $H_n$ pulls back to the cycles and extends to $C_n$; the extension vanishes on boundaries, so gives a cocycle mapping to that functional. If a cocycle vanishes on cycles, the rule $b(\mathrm{\partial }c)=f(c)$ is well-defined on the boundary subspace in degree $n-1$ and extends to $C_{n-1}$, giving $f=\delta b$ after that extension. These are exactly the surjectivity and injectivity arguments in [F3]; both use its AC extension clause. Conversely every coboundary vanishes on cycles by step 1.2. Thus evaluation is the asserted natural isomorphism. It is not an alternative definition, and no global family of cochain representatives was chosen.

If $X=\mathrm{\varnothing }$, all chain and cochain groups are zero and evaluation is the unique isomorphism between zero spaces in every degree. If $X$ is a point, there is one simplex in every nonnegative degree and ${\mathrm{\partial }}_k$ is multiplication by ${\sum }_{i=0}^k(-1{)}^i$ for $k>0$: it is the identity in positive even degrees and zero in odd degrees, with ${\mathrm{\partial }}_0=0$. Therefore $H_0=\mathbb{R}$ and $H_k=0$ for $k>0$; dually $H^0=\mathbb{R}$ and $H^k=0$ for $k>0$. Evaluation in degree zero sends the constant scalar cochain a to the functional $r\mapsto ar$, an isomorphism without choice. In negative degrees both sides vanish. For general X at degree zero there is no incoming coboundary, so the injectivity argument uses no negative-degree extension. Constant and repeated simplices are retained in these unnormalized complexes; the representative computations in step 1.2 apply to them as written.