Lemma 14.26.10. Let $\mathcal{C}$ be a category. Let $T$ be a set. For $t \in T$ let $X_ t$, $Y_ t$ be simplicial objects of $\mathcal{C}$. Assume $X = \prod _{t \in T} X_ t$ and $Y = \prod _{t \in T} Y_ t$ exist.

If $X_ t$ and $Y_ t$ are homotopy equivalent for all $t \in T$ and $T$ is finite, then $X$ and $Y$ are homotopy equivalent.

For $t \in T$ let $a_ t, b_ t : X_ t \to Y_ t$ be morphisms. Set $a = \prod a_ t : X \to Y$ and $b = \prod b_ t : X \to Y$.

If there exists a homotopy from $a_ t$ to $b_ t$ for all $t \in T$, then there exists a homotopy from $a$ to $b$.

If $T$ is finite and $a_ t, b_ t : X_ t \to Y_ t$ for $t \in T$ are homotopic, then $a$ and $b$ are homotopic.

**Proof.**
If $h_ t = (h_{t, n , i})$ is a homotopy from $a_ t$ to $b_ t$ (see Remark 14.26.4), then $h = (\prod _ t h_{t, n, i})$ is a homotopy from $\prod a_ t$ to $\prod b_ t$. This proves (2).

Proof of (3). Choose $t \in T$. There exists an integer $n \geq 0$ and a chain $a_ t = a_{t, 0}, a_{t, 1}, \ldots , a_{t, n} = b_ t$ such that for every $1 \leq i \leq n$ either there is a homotopy from $a_{t, i - 1}$ to $a_{t, i}$ or there is a homotopy from $a_{t, i}$ to $a_{t, i - 1}$. If $n = 0$, then we pick another $t$. (We're done if $a_ t = b_ t$ for all $t \in T$.) So assume $n > 0$. By Example 14.26.3 there are is a homotopy from $b_{t'}$ to $b_{t'}$ for all $t' \in T \setminus \{ t\} $. Thus by (2) there is a homotopy from $a_{t, n - 1} \times \prod _{t'} b_{t'}$ to $b$ or there is a homotopy from $b$ to $a_{t, n - 1} \times \prod _{t'} b_{t'}$. In this way we can decrease $n$ by $1$. This proves (3).

Part (1) follows from part (3) and the definitions.
$\square$

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