Proof. Since the C-removal lemma is not polynomial, there is a function $\delta\,:\,(0,1) \rightarrow (0,1)$ such that $1/\delta(\varepsilon)$ grows faster than any polynomial in $1/\varepsilon$ , and such that for every $\varepsilon \gt 0$ and large enough n there is an n-vertex k-graph $H_1$ which contains a collection $\mathcal{C}$ of $\varepsilon n^k$ edge-disjoint copies of C but only $\delta n^{v(C)}$ copies of C altogether. Let H be the v(F)-blowup of $H_1$ . Note that the v(F)-blowup of C contains a copy of F. Also, copies of F corresponding to different copies of C from $\mathcal{C}$ are edge-disjoint. Hence, H has a collection of $\varepsilon n^k = \varepsilon (v(H)/v(F))^k = \Omega(\varepsilon \cdot v(H)^k) = \varepsilon' v(H)^k$ edge-disjoint copies of F, for a suitable $\varepsilon' = \Omega(\varepsilon)$ . Let us bound the total number of copies of F in H. Since C is a subgraph of F, each copy of F must contain a copy of C. Let $\varphi\,:\,V(H) \rightarrow V(H_1)$ be the natural homomorphism from H to $H_1$ (as defined above). For each copy $C'$ of C in H, consider the subgraph $\varphi(C')$ of $H_1$ . The number of copies $C'$ of C with $v(\varphi(C')) \lt v(C)$ is at most $v(F)^{v(C)} \cdot O(n^{v(C) - 1}) \leq \delta n^{v(C)}$ , provided that n is large enough. The number of copies $C'$ of C with $\varphi(C') \cong C$ is at most $v(F)^{v(C)} \cdot \delta n^{v(C)} = O(\delta n^{v(C)})$ , because $H_1$ contains at most $\delta n^{v(C)}$ copies of C. So in total, H contains at most $O(\delta n^{v(C)})$ copies of C. This means that H contains at most $O(\delta n^{v(C)}) \cdot v(H)^{v(F) - v(C)} = {}O(\delta \cdot v(H)^{v(F)}) = \delta' v(H)^{v(F)}$ copies of F, for a suitable $\delta' = O(\delta)$ . Note that $1/\delta'$ is super-polynomial in $1/\varepsilon'$ . This shows that the F-removal lemma is not polynomial.

Proof. We first observe that every homomorphism from a core F to itself is an isomorphism. Indeed, by definition, F is the core of itself, meaning that there is no homomorphism from F to a subgraph $F_0$ of F with $V(F_0) \subsetneq V(F)$ . Hence, every homomorphism from F to itself is a bijection, and hence an isomorphism. The assertion of the claim now follows from the fact that $\varphi_{|{V(F')}}$ is a homomorphism from $F'$ (which is a copy of F) to F.

Proof. We construct the family $\mathcal{F}$ as follows. For each $S \in \mathcal{S}$ and each r-tuple $A \in V_{s+1} \times \ldots \times V_{s+r}$ , add $S \cup A$ to $\mathcal{F}$ with probability $1/(Cn^{r-k+\ell})$ and independently, where C is a large constant to be chosen later. (i) is satisfied by definition. Let us estimate the number of pairs $F_1,F_2 \in \mathcal{F}$ violating (iii); denote this number by B. We claim that

(2.1) \begin{equation} {} {}\mathbb{E}[B] = O_{s,r,k}\left(\frac{1}{C^2} \right) \cdot |\mathcal{S}| \cdot n^{k-\ell}. {}\end{equation}

To this end, suppose that $F_1,F_2 \in \mathcal{F}$ violate (iii), and write $F_1 = S_1 \cup A_1$ and $F_2 = S_2 \cup A_2$ , where $S_1,S_2 \in \mathcal{F}$ and $A_1,A_2 \in V_{s+1} \times \ldots \times V_{s+r}$ . Suppose first that $S_1 = S_2$ . Then there are $|\mathcal{S}|$ choices for $S_1,S_2$ . Also, to violate (iii), it must hold that $|A_1 \cap A_2| \geq k-\ell$ . The number of choices of $A_1,A_2 \in V_{s+1} \times \ldots \times V_{s+r}$ with $|A_1 \cap A_2| \geq k-\ell$ is at most $n^{r} \cdot \binom{r}{k-\ell} \cdot n^{r-k+\ell}$ . Finally, the probability that $F_1,F_2 \in \mathcal{F}$ is $1/(Cn^{r-k+\ell})^2$ . Hence, the expected number of violations of this type (i.e., with $S_1 = S_2$ ) is at most $|\mathcal{S}| \cdot n^{r} \cdot \binom{r}{k-\ell} \cdot n^{r-k+\ell} \cdot 1/(Cn^{r-k+\ell})^2 = O_{s,r,k}\left({1}/{C^2} \right) \cdot |\mathcal{S}| \cdot n^{k-\ell}$ .

Now consider the case that $S_1 \neq S_2$ , and put $t := |S_1 \cap S_2|$ . As the sets in $\mathcal{S}$ are pairwise $\ell$ -disjoint, we have $t \leq \ell-1$ . Also, the number of choices for $S_1,S_2 \in \mathcal{S}$ with $|S_1 \cap S_2| = t$ is at most $|\mathcal{S}| \cdot \binom{s}{t} \cdot n^{\ell-t}$ , again using that the sets in $\mathcal{S}$ are pairwise $\ell$ -disjoint. In order for $F_1,F_2$ to violate (iii), we must have $|A_1 \cap A_2| \geq k-t$ . The number of choices for $A_1,A_2 \in V_{s+1} \times \ldots \times V_{s+r}$ with $|A_1 \cap A_2| \geq k-t$ is at most $n^{r} \cdot \binom{r}{k-t} \cdot n^{r-k+t}$ . Finally, as before, the probability that $F_1,F_2 \in \mathcal{F}$ is $1/(Cn^{r-k+\ell})^2$ . Hence, the expected number of violations of this type (i.e., with $S_1 \neq S_2$ ) is at most

$$ {}\sum_{t=0}^{\ell-1} \left[ |\mathcal{S}| \cdot \binom{s}{t} \cdot n^{\ell-t} \cdot n^{r} \cdot \binom{r}{k-t} \cdot n^{r-k+t} \cdot \left(\frac{1}{Cn^{r-k+\ell}}\right)^2 \right] = {}O_{s,r,k}\left(\frac{1}{C^2} \right) \cdot |\mathcal{S}| \cdot n^{k-\ell}. $$

This proves (2 $ \cdot $ 1). Now note that the expected size of $\mathcal{F}$ is $|\mathcal{S}| \cdot n^r \cdot {1}/{Cn^{r-k+\ell}} = {1}/{C} \cdot |\mathcal{S}| \cdot n^{k-\ell}$ . So by choosing C to be large enough (as a function of s,r,k), we can guarantee that $\mathbb{E}[|\mathcal{F}| - B] \geq {1}/{2C} \cdot |\mathcal{S}| \cdot n^{k-\ell}$ . By fixing such a choice of $\mathcal{F}$ and deleting one set $F \in \mathcal{F}$ from each violation, we get the required conclusion.

Proof. Apply Lemma 2·3 with $s = \ell = 0$ and $\mathcal{S} = \{\emptyset\}$ .

Proof. Let $\varepsilon \gt 0$ and let n be large enough. Let m be the largest integer satisfying $m^{\delta} \leq 1/\varepsilon$ , so that $m \geq (1/\varepsilon)^{1/(2\delta)}$ , say. Let H be the k-graph guaranteed to exist by the assumption of the lemma, but with m in place of n. So H has m vertices, is homomorphic to F, contains a collection $\mathcal{F}$ of $m^{k-\delta} \geq \varepsilon m^k$ edge-disjoint copies of F, but has at most $m^{v(F) - 1}$ copies of F altogether.

Let G be the ${n}/{m}$ -blowup of H. Each $F' \in \mathcal{F}$ gives rise to $\Omega(({n}/{m})^k)$ k-disjoint (and hence also edge-disjoint) copies of F in G, by Lemma 2·4 applied with $r = v(F)$ and with ${n}/{m}$ in place of n. Copies arising from different $F'_1,F'_2 \in \mathcal{F}$ are edge-disjoint, because the copies in $\mathcal{F}$ are edge-disjoint. Altogether, this gives a collection of $\varepsilon m^k \cdot \Omega(({n}/{m})^k) = \Omega(\varepsilon n^k)$ edge-disjoint copies of F in G.

Let us upper-bound the total number of copies of F in G. By assumption, there is a homomorphism $\varphi$ from H to F. Let $\psi$ be the “natural” homomorphism from G to H (as described in the beginning of this section). Then $\varphi \circ \psi$ is a homomorphism from G to F. By Claim 2·2, for every copy $F'$ of F in G the map $(\varphi \circ \psi)|_{{V(F')}}$ is an isomorphism from $F'$ to F. We claim that this means that $\psi$ maps every copy $F'$ of F in G onto a copy of F in H. Indeed, $\psi|_{{V(F')}}$ must be injective (otherwise $(\varphi \circ \psi)|_{{V(F')}}$ would not be an isomorphism), and since $\psi|_{{V(F')}}$ must map edges to edges (on account of being a homomorphism) its image must contain a copy of F. We thus see that every copy of F in G must come from the blown-up copies of F in H. But each copy of F in H gives rise to $({n}/{m})^{v(F)}$ copies of F in G. Hence, the total number of copies of F in G is at most

$$ m^{v(F) - 1} \cdot (n/m)^{v(F)} = n^{v(F)}/m \leq \varepsilon^{1/(2\delta)} \cdot n^{v(F)}.$$

Since $\delta \gt 0$ is arbitrary, this shows that the F-removal lemma is not polynomial.

Proof. Suppose that the vertices of $C_{t}$ are $1,2,\ldots,t$ (appearing in this order along the cycle). Take a set $B \subseteq [n/t]$ , $|B| \geq n/e^{O\sqrt{\log n}}$ , with no non-trivial solution to the linear equation $y_1 + \ldots + y_{t-1} = (t-1)y_{t}$ with $y_1,\ldots,y_{t} \in B$ (where a solution is trivial if $y_1=y_2=\cdots=y_{t}$ ). The existence of such a set B is by a simple generalisation of Behrend’s construction [ Reference Behrend3 ] of sets avoiding 3-term arithmetic progressions, see [Reference Alon1, lemma 3·1]. Take pairwise-disjoint sets $V_1,\ldots,V_{t}$ of size n each, and identify each $V_i$ with [n]. For each $x \in [n/t]$ and $y \in B$ , add to G a canonical copy $S_{x,y}$ of $C_{t}$ on the vertices $v_i = x + (i - 1)y \in V_i$ , $i = 1,\ldots,t$ . Note that $x + (i - 1)y \leq x + (t-1)y \leq n$ , so $v_i$ indeed “fits” into $V_i = [n]$ . The copies $S_{x,y}$ (where $x \in [n/t], y \in B$ ) are 2-disjoint. Indeed, if $S_{x_1,y_1},S_{x_2,y_2}$ intersect in $V_i$ and in $V_j$ , then $x_1 + (i - 1)y_1 = x_2 + (i - 1)y_2$ and $x_1 + (j - 1)y_1 = x_2 + (j - 1)y_2$ , and solving this system of equations gives $x_1 = x_2, y_1 = y_2$ . The number of copies $S_{x,y}$ is ${n}/{t} \cdot |B| \geq n^2/e^{O\sqrt{\log n}}$ .

Let us bound the total number of canonical copies of $C_{t}$ in G. Fix a canonical copy with vertices $v_1,\ldots,v_{t}$ , $v_i \in V_i$ . For $1 \leq j \leq t-1$ , let $x_j \in [n/t], y_j \in B$ be such that $v_{j},v_{j+1} \in S_{x_j,y_j}$ . Similarly, let $x_{t} \in [n/t], y_{t} \in B$ such that $v_1,v_{t} \in S_{x_{t},y_{t}}$ . Then we have $v_{j+1} - v_{j} = y_j$ for every $1 \leq j \leq t-1$ , and $v_{t} - v_{1} = (t-1)y_{t}$ . So $y_1 + \ldots + y_{t-1} = (t-1)y_{t}$ . By our choice of B, we have $y_1 = \ldots = y_{t} =: y$ . Now, for each $1 \leq j \leq t-1$ we have $x_j = v_{j+1} - j \cdot y = x_{j+1}$ , so $x_1 = \ldots = x_{t} =: x$ . So we see that the only canonical copies of $C_{t}$ in G are the copies $S_{x,y}$ . Their number is at most $n^2 \leq n^{t-1}$ , as required.

Proof. Take a set $B \subseteq [n/s]$ , $|B| \geq n/e^{O\sqrt{\log n}}$ , with no non-trivial solution to $y_1 + y_2 = 2y_3$ , $y_1,y_2,y_3 \in B$ . Take pairwise-disjoint sets $V_1,\ldots,V_s$ of size n each, and identify each $V_i$ with [n]. For each $x_1,\ldots,x_{s-2} \in [n/s]$ and $y \in B$ , add to G a copy $K_{x_1,\ldots,x_{s-2},y}$ of $K_s^{(s-1)}$ on the vertices

$$ x_1 \in V_1, \,\,\,x_2 \in V_2, \,\,\,\ldots \,\,\,x_{s-2} \in V_{s-2}, \,\,\,y+\sum^{s-2}_{i=1}x_i \in V_{s-1}, \,\,\,2y+\sum^{s-2}_{i=1}x_i \in V_s.$$

It is easy to see that these copies are $(s-1)$ -disjoint, because fixing any $s-1$ of the s coordinates allows to solve for $x_1,\ldots,x_{s-2},y$ . Also, the number of copies thus placed is $(n/s)^{s-2} \cdot |B| \geq n^{s-1}/e^{O\sqrt{\log n}}$ . Let us show that there are no other copies of $K_s^{(s-1)}$ in G. This would imply that the total number of copies of $K_s^{(s-1)}$ in G is $(n/s)^{s-2} \cdot |B| \leq n^{s-1}$ . So suppose that $v_1 \in V_1,\ldots,v_s \in V_s$ form a copy of $K_s^{(s-1)}$ . Let $x^{(i)} = (x^{(i)}_1,\ldots,x^{(i)}_{s-2}) \in [n/s]^{s-2}$ and $y_i \in B$ , $i = 1,2,3$ , be such that $\{v_2,\ldots,v_s\} \in K_{x^{(1)},y_1}$ , $\{v_1,\ldots,v_{s-1}\} \in K_{x^{(2)},y_2}$ and $\{v_1,\ldots,v_{s-2},v_s\} \in K_{x^{(3)},y_3}$ . Then $x^{(2)}_1 = x^{(3)}_1 = v_1$ and

(2.2) \begin{equation} {} {}x^{(1)}_j = x^{(2)}_j = x^{(3)}_j = v_j \text{ for every } 2 \leq j \leq s-2. {} {}\end{equation}

Also, $v_s - v_{s-1} = y_1$ , $v_{s-1} - v_1 = x^{(2)}_2 + \cdots + x^{(2)}_{s-2} + y_2$ and $v_s - v_1 = x^{(3)}_2 + \cdots + x^{(3)}_{s-2} + 2y_3$ . Combining these three equations and using (2 $ \cdot $ 2), we get $y_1 + y_2 = 2y_3$ , and so $y_1 = y_2 = y_3 =: y$ by our choice of B. Also, $x^{(1)}_1 = v_{s-1} - (v_2 + \cdots + v_{s-2} + y) = x^{(2)}_1$ . So $x^{(1)} = x^{(2)} = x^{(3)}$ .

Proof. It will be convenient to write $|V(F)| = t+r$ and assume that $V(F) = [t+r]$ , where $(1,2,\ldots,t,1)$ is an induced cycle in $\partial_2 F$ and $t \geq 4$ . It follows that $|e \cap \{1,\ldots,t\}| \leq 2$ for every $e \in E(F)$ . Take disjoint sets $V_1,\ldots,V_{t+r}$ of size n each. Let G be the t-partite graph with sides $V_1,\ldots,V_t$ given by Lemma 2·6. Let $\mathcal{S}$ be a collection of $n^2/e^{O(\sqrt{\log n})}$ 2-disjoint canonical copies of $C_t$ in G. Apply Lemma 2·3 toFootnote ⁴ $\mathcal{S}$ with $s=t$ and $\ell = 2$ to obtain a family $\mathcal{F} \subseteq V_1 \times \ldots \times V_{t+r}$ satisfying Items 1-3 in that lemma. Note that $r \geq k-2 = k-\ell$ , because each edge of F contains at most two vertices from $\{1,\ldots,t\}$ and hence at least $k-2$ vertices from $\{t+1,\ldots,t+r\}$ . Therefore, the conditions of Lemma 2·3 are satisfied. Define the hypergraph H by placing a canonical copy of F on each $F' \in \mathcal{F}$ . We claim that these copies of F are edge-disjoint. Indeed, suppose by contradiction that the copies on $F_1,F_2 \in \mathcal{F}$ share an edge e. Then $|F_1 \cap F_2| \geq k$ . By Lemma 2·3(iii), we have $\#\{t+1 \leq i \leq t+r\,:\,F_1(i) = F_2(i)\} \leq k-3$ . This implies that $\#\{1 \leq i \leq t\,:\,e \cap V_i \neq \emptyset\} \geq 3$ . But this means that in F there is an edge which intersects $\{1,\ldots,t\}$ in at least 3 vertices, a contradiction. So the F-copies in $\mathcal{F}$ are indeed edge-disjoint. Their number is $|\mathcal{F}| \geq \Omega(|\mathcal{S}|n^{k-2}) \geq n^k/e^{O(\sqrt{\log n})}$ , by Lemma 2·3(ii).

To complete the proof, it remains to show that H has at most $n^{t+r-1}$ copies of F. Observe that H is homomorphic to F; indeed, the map $\varphi$ which sends $V_j \mapsto j$ , $j = 1,\ldots,t+r$ , is such a homomorphism. Let $F^*$ be a copy of F in H. Since F is a core and $\varphi$ is a homomorphism from H to F, we can apply Claim 2·2 to conclude that $F^*$ must have the form $v_1,\ldots,v_{t+r}$ , with $v_i \in V_i$ playing the role of i for each $i = 1,\ldots,t+r$ . We claim that $v_1,\ldots,v_t$ form a canonical copy of $C_t$ in Footnote ⁵ G. To see this, fix any $1 \leq i \leq t$ and let us show that $\{v_i,v_{i+1}\} \in E(G)$ , with indices taken modulo t. Since $\{i,i+1\}$ is an edge of $\partial_2 F$ , there must be an edge $e \in E(F)$ containing $i,i+1$ . Then $\{v_a\,:\,a \in e\} \in E(F^*) \subseteq E(H) = \bigcup_{F' \in \mathcal{F}}{E(F')}$ . Let $F' \in \mathcal{F}$ such that $\{v_a\,:\,a \in e\} \in E(F')$ . By Lemma 2·3(i), we have $S' := F'|_{V_1 \times \ldots \times V_t} \in \mathcal{S}$ . Now, $S'$ is the vertex set of a canonical copy of $C_t$ in G, and hence $\{v_i,v_{i+1}\} \in E(G)$ , as required. This proves our claim that $v_1,\ldots,v_t$ form a canonical copy of $C_t$ in G. Summarising, every copy of F in H contains the vertices of a canonical copy of $C_t$ in G. By the guarantees of Lemma 2·6, the number of canonical copies of $C_t$ in G is at most $n^{t-1}$ . Hence, the number of copies of F in H is at most $n^{t-1} \cdot n^{r} = n^{t+r-1}$ , as required.

Proof. The proof is very similar to that of Lemma 2·8. Assume that $I = [s]$ , $V(F) = [s+r]$ . Take disjoint sets $V_1,\ldots,V_{s+r}$ of size n each. Let G be the s-partite $(s-1)$ -graph with sides $V_1,\ldots,V_s$ given by Lemma 2·7. Let $\mathcal{S}$ be a collection of $n^{s-1}/e^{O(\sqrt{\log n})}$ $(s-1)$ -disjoint copies of $K_{s}^{(s-1)}$ in G. Apply Lemma 2·3 to $\mathcal{S}$ with $\ell = s-1$ to obtain a family $\mathcal{F} \subseteq V_1 \times \ldots \times V_{s+r}$ satisfying (i)-(iii) in that lemma. Define the hypergraph H by placing a canonical copy of F on each $F' \in \mathcal{F}$ . These copies of F are edge-disjoint. Indeed, suppose by contradiction that the copies on $F_1,F_2 \in \mathcal{F}$ share an edge e. Then $|F_1 \cap F_2| \geq k$ , and hence $\#\{s+1 \leq i \leq s+r\,:\,F_1(i) = F_2(i)\} \leq k-\ell-1 = k-s$ by Lemma 2·3(iii). But then $\#\{1 \leq i \leq s\,:\,e \cap V_i \neq \emptyset\} = s$ , meaning that there is an edge of F which contains $I = [s]$ , a contradiction to the assumption of the lemma. So the F-copies in $\mathcal{F}$ are indeed edge-disjoint. Also, $|\mathcal{F}| \geq \Omega(|\mathcal{S}|n^{k-s+1}) \geq n^k/e^{O(\sqrt{\log n})}$ , using Lemma 2·3(ii).

The map $V_j \mapsto j$ , $j = 1,\ldots,s+r$ is a homomorphism from H to F. Let us bound the number of copies of F in H. By Claim 2·2, every copy $F^*$ of F must be of the form $v_1,\ldots,v_{s+r}$ , with $v_i \in V_i$ playing the role of i for each $i = 1,\ldots,s+r$ . We claim that $v_1,\ldots,v_s$ span a copy of $K_s^{(s-1)}$ in G. So let $J \in \binom{[s]}{s-1}$ . Since $(\partial_{s-1} F)[I] \cong K_s^{(s-1)}$ , there is an edge $e \in E(F)$ with $J \subseteq e$ . Since $F^*$ is a canonical copy of F, we have $\{v_i\,:\,i \in e\} \in E(F^*) \subseteq E(H) = \bigcup_{F' \in \mathcal{F}}{E(F')}$ . Let $F' \in \mathcal{F}$ be such that $\{v_i\,:\,i \in e\} \in E(F')$ . By Lemma 2·3(i), we have $S' := F'|_{V_1 \times \ldots \times V_s} \in \mathcal{S}$ . Now, $S'$ is a canonical copy of $K_s^{(s-1)}$ in G, and hence $\{v_i\,:\,i \in J\} \in E(G)$ , as required. So we see that every copy of F in H contains the vertices of a copy of $K_s^{(s-1)}$ in G. By the guarantees of Lemma 2·6, G has at most $n^{s-1}$ copies of $K_s^{(s-1)}$ . Hence, H has at most $n^{s-1} \cdot n^{r} = n^{s+r-1}$ copies of F, as required.

Proof of Lemma 1·2. Let X be a clique of size $k+1$ in $\partial_2 F$ . Let I be a smallest subset of X which is not contained in an edge of F. Note that I is well-defined (because X itself is not contained in any edge of F, as $|X| = k+1$ ). Also, $|I| \geq 3$ because every pair of vertices in X is contained in some edge, as X is a clique in $\partial_2F$ . Put $s = |I|$ . Then $(\partial_{s-1}F)[I] \cong K_s^{(s-1)}$ and $|e \cap I| \leq s-1$ for every $e \in E(F)$ , by the choice of I. Now the assertion of Lemma 1·2 follows by combining Lemmas 2·5 and 2·9.