In CGWH, is every cofibration an inclusion with closed image?

Question

As the title suggests, in CGWH, is every cofibration an inclusion with closed image?

Answer 1

The main obstruction in the CGWH case (as opposed to the Hausdorff case) is in proving that retracts have closed image. But if $r: X \to X$ is idempotent for a compactly generated space $X$, then$$r(X) = \{x : x = r(x)\} = (r \times \text{Id})^{-1} \Delta X,$$which is closed by continuity and the following lemma in $\S6.1$ of May's A Concise Course in Algebraic Topology.

Lemma. If $i: A \to X$ is a cofibration and $g: A \to B$ is any map, then the induced map $B \to B \cup_g X$ is a cofibration.

Basically, it is all shuffled into the $k$-fication of products.

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Update. In more detail, suppose $i: A \to X$ is a cofibration. By the homotopy extension property, the map $r: X \times I \to Mi$ described on page 44 of May's A Concise Course in Algebra Topology exists, along with $j: Mi \to X \times I$ such that $r \circ j = \text{Id}$. This implies $j$ is a monomorphism. Since $j|_{A \times I} = i \times \text{Id}$, it follows that$$i(a) = i(b) \implies (i \times \text{Id})(a, 0) = (i \times \text{Id})(b, 0) \implies a = b.$$Thus, $i$ is injective.

We claim $j(Mi) \subset X \times I$ is closed. To see this, let$$S = \{(x, t) \in X \times I : j \circ r(x, t) = (x, t)\}.$$Then since$$j \circ r \circ j = j \circ \text{Id} = j,$$we see that $j(Mi) \subset S$. Conversely, if $(x, t) \in S$, then$$(x, t) = j \circ r(x, t) \in j(Mi),$$so $S \subset j(Mi)$. Thus, it suffices to prove $S$ is closed. This follows from the following lemma.

Lemma. Suppose $Y$ is compactly generated. Let $f: Y \to Y$ be a map. Then $$\text{Fix}(f) = \{y \in Y : f(y) = y\}$$is closed in $Y$.

Proof. Let $\Delta Y = \{(y, z) \in Y \times Y : y = z\}$. Consider the following lemma in $\S6.1$ of May's A Concise Course in Algebra Topology.

Lemma. If $X$ is a $k$-space, then $X$ is weak Hausdorff if and only if $\Delta X$ is closed in $X \times X$.

By this lemma, we know $\Delta Y$ is closed in $Y \times Y$. Consider the map $\phi: Y \to Y \times Y$ given by $y \mapsto (y, f(y))$. This is continuous, so $\text{Fix}(f) = \phi^{-1}(\Delta Y)$ is closed in $Y$.$$\tag*{$\square$}$$Since $X \times \{1\} \subset X \times I$ is closed, we see$$j(Mi) \cap (X \times \{1\}) = (i \times \text{Id})(A \times \{1\}) = i(A) \times \{1\}$$is closed in $X \times I$. Let $\pi: X \times I \to X$ be the projection. We claim this is a closed map.

Lemma. If $Y$ is compact, the projection $\pi: X \times Y \to X$ is a closed map.

Proof. Let $S \subset X \times Y$ be closed. Choose $x \in X \setminus \pi(S)$. For any $y \in Y$, we know $(x, y) \notin S$, so there exists a neighborhood $U_{(x, y)} \times Y_{(x, y)} \subset S^c$ containing $(x, y)$. The $V_{(x, y)}$ cover $Y$, so there is some finite subcover $V_{(x, y_1)}, \dots, V_{(x, y_n)}$. Let $U = \bigcap_{k=1}^n U_{(x, y_k)}$. Then $U \subset \pi(S)^c$ is an open neighborhood of $x$. Thus, $\pi(S)^c$ is open, i.e. $\pi(S)$ is closed.$$\tag*{$\square$}$$Thus, we conclude $\pi(i(A) \times \{1\}) = i(A)$ is closed in $X$.

Lastly, we claim $i: A \to X$ is a homeomorphism onto its image. To see this, let $h_t: A \to Mi$ be the map $h_t(A) = (a, t)$. Note that $h_0(a) = (a, 0)$ so $h_0$ restricts the inclusion $\nu: X \hookrightarrow Mi$. Thus, by the homotopy extension property, we get a homotopy $\widetilde{h}_t: X \to Mi$ with $\widetilde{h}_t \circ i = h_t$. Since $h_1$ is a homeomorphism $A \to A \times \{1\}$, we have $h_1^{-1} \circ \widetilde{h}_1 \circ i = \text{Id}$. Thus, $h_1^{-1} \circ \widetilde{h}_1 : i(A) \to A$ is a continuous inverse for $i$.

Answer 2

Yes, see Lemma 2.4.5 and Corollary 2.4.6 in Hovey's book Model Categories. This is in the category of all spaces. It's also true for CGWH as is stated in a proof at the top of page 59 of Hovey's book.